diff --git a/README.md b/README.md index 358cad4..6599c95 100644 --- a/README.md +++ b/README.md @@ -16,6 +16,26 @@ This means you'll always find an AI-generated solution to compare with or learn # Problems we've attempted so far:
+Problems from December, 2025 + +| πŸ“† Date | βš™οΈ Problem | πŸ“ Link to notes | 🚦 Difficulty | +|--------------|-------------------------------------------------------------------------------------------------------------|-----------------------------------------------------------------------------------------------------------|------------| +| December 1, 2025 | [2141](https://leetcode.com/problems/maximum-running-time-of-n-computers/description/?envType=daily-question) | [Click here](https://github.com/ContextLab/leetcode-solutions/tree/main/problems/2141) | πŸ”΄ Hard | +| December 2, 2025 | [3623](https://leetcode.com/problems/count-number-of-trapezoids-i/description/?envType=daily-question) | [Click here](https://github.com/ContextLab/leetcode-solutions/tree/main/problems/3623) | 🟑 Medium | +| December 3, 2025 | [3625](https://leetcode.com/problems/count-number-of-trapezoids-ii/description/?envType=daily-question) | [Click here](https://github.com/ContextLab/leetcode-solutions/tree/main/problems/3625) | πŸ”΄ Hard | +| December 4, 2025 | [2211](https://leetcode.com/problems/count-collisions-on-a-road/description/?envType=daily-question) | [Click here](https://github.com/ContextLab/leetcode-solutions/tree/main/problems/2211) | 🟑 Medium | +| December 5, 2025 | [3432](https://leetcode.com/problems/count-partitions-with-even-sum-difference/description/?envType=daily-question) | [Click here](https://github.com/ContextLab/leetcode-solutions/tree/main/problems/3432) | 🟒 Easy | +| December 6, 2025 | [3578](https://leetcode.com/problems/count-partitions-with-max-min-difference-at-most-k/description/?envType=daily-question) | [Click here](https://github.com/ContextLab/leetcode-solutions/tree/main/problems/3578) | 🟑 Medium | +| December 7, 2025 | [1523](https://leetcode.com/problems/count-odd-numbers-in-an-interval-range/description/?envType=daily-question) | [Click here](https://github.com/ContextLab/leetcode-solutions/tree/main/problems/1523) | 🟒 Easy | +| December 8, 2025 | [1925](https://leetcode.com/problems/count-square-sum-triples/description/?envType=daily-question) | [Click here](https://github.com/ContextLab/leetcode-solutions/tree/main/problems/1925) | 🟒 Easy | +| December 9, 2025 | [3583](https://leetcode.com/problems/count-special-triplets/description/?envType=daily-question) | [Click here](https://github.com/ContextLab/leetcode-solutions/tree/main/problems/3583) | 🟑 Medium | +| December 10, 2025 | [3577](https://leetcode.com/problems/count-the-number-of-computer-unlocking-permutations/description/?envType=daily-question) | [Click here](https://github.com/ContextLab/leetcode-solutions/tree/main/problems/3577) | 🟑 Medium | +| December 11, 2025 | [3531](https://leetcode.com/problems/count-covered-buildings/description/?envType=daily-question) | [Click here](https://github.com/ContextLab/leetcode-solutions/tree/main/problems/3531) | 🟑 Medium | +| December 12, 2025 | [3433](https://leetcode.com/problems/count-mentions-per-user/description/?envType=daily-question) | [Click here](https://github.com/ContextLab/leetcode-solutions/tree/main/problems/3433) | 🟑 Medium | + +
+ +
Problems from November, 2025 | πŸ“† Date | βš™οΈ Problem | πŸ“ Link to notes | 🚦 Difficulty | @@ -26,6 +46,30 @@ This means you'll always find an AI-generated solution to compare with or learn | November 4, 2025 | [3318](https://leetcode.com/problems/find-x-sum-of-all-k-long-subarrays-i/description/?envType=daily-question) | [Click here](https://github.com/ContextLab/leetcode-solutions/tree/main/problems/3318) | 🟒 Easy | | November 5, 2025 | [3321](https://leetcode.com/problems/find-x-sum-of-all-k-long-subarrays-ii/description/?envType=daily-question) | [Click here](https://github.com/ContextLab/leetcode-solutions/tree/main/problems/3321) | πŸ”΄ Hard | | November 6, 2025 | [3607](https://leetcode.com/problems/power-grid-maintenance/description/?envType=daily-question) | [Click here](https://github.com/ContextLab/leetcode-solutions/tree/main/problems/3607) | 🟑 Medium | +| November 7, 2025 | [2528](https://leetcode.com/problems/maximize-the-minimum-powered-city/description/?envType=daily-question) | [Click here](https://github.com/ContextLab/leetcode-solutions/tree/main/problems/2528) | πŸ”΄ Hard | +| November 8, 2025 | [1611](https://leetcode.com/problems/minimum-one-bit-operations-to-make-integers-zero/description/?envType=daily-question) | [Click here](https://github.com/ContextLab/leetcode-solutions/tree/main/problems/1611) | πŸ”΄ Hard | +| November 9, 2025 | [2169](https://leetcode.com/problems/count-operations-to-obtain-zero/description/?envType=daily-question) | [Click here](https://github.com/ContextLab/leetcode-solutions/tree/main/problems/2169) | 🟒 Easy | +| November 10, 2025 | [3542](https://leetcode.com/problems/minimum-operations-to-convert-all-elements-to-zero/description/?envType=daily-question) | [Click here](https://github.com/ContextLab/leetcode-solutions/tree/main/problems/3542) | 🟑 Medium | +| November 11, 2025 | [474](https://leetcode.com/problems/ones-and-zeroes/description/?envType=daily-question) | [Click here](https://github.com/ContextLab/leetcode-solutions/tree/main/problems/474) | 🟑 Medium | +| November 12, 2025 | [2654](https://leetcode.com/problems/minimum-number-of-operations-to-make-all-array-elements-equal-to-1/description/?envType=daily-question) | [Click here](https://github.com/ContextLab/leetcode-solutions/tree/main/problems/2654) | 🟑 Medium | +| November 13, 2025 | [3228](https://leetcode.com/problems/maximum-number-of-operations-to-move-ones-to-the-end/description/?envType=daily-question) | [Click here](https://github.com/ContextLab/leetcode-solutions/tree/main/problems/3228) | 🟑 Medium | +| November 14, 2025 | [2536](https://leetcode.com/problems/increment-submatrices-by-one/description/?envType=daily-question) | [Click here](https://github.com/ContextLab/leetcode-solutions/tree/main/problems/2536) | 🟑 Medium | +| November 15, 2025 | [3234](https://leetcode.com/problems/count-the-number-of-substrings-with-dominant-ones/description/?envType=daily-question) | [Click here](https://github.com/ContextLab/leetcode-solutions/tree/main/problems/3234) | 🟑 Medium | +| November 16, 2025 | [1513](https://leetcode.com/problems/number-of-substrings-with-only-1s/description/?envType=daily-question) | [Click here](https://github.com/ContextLab/leetcode-solutions/tree/main/problems/1513) | 🟑 Medium | +| November 17, 2025 | [1437](https://leetcode.com/problems/check-if-all-1s-are-at-least-length-k-places-away/description/?envType=daily-question) | [Click here](https://github.com/ContextLab/leetcode-solutions/tree/main/problems/1437) | 🟒 Easy | +| November 18, 2025 | [717](https://leetcode.com/problems/1-bit-and-2-bit-characters/description/?envType=daily-question) | [Click here](https://github.com/ContextLab/leetcode-solutions/tree/main/problems/717) | 🟒 Easy | +| November 19, 2025 | [2154](https://leetcode.com/problems/keep-multiplying-found-values-by-two/description/?envType=daily-question) | [Click here](https://github.com/ContextLab/leetcode-solutions/tree/main/problems/2154) | 🟒 Easy | +| November 20, 2025 | [757](https://leetcode.com/problems/set-intersection-size-at-least-two/description/?envType=daily-question) | [Click here](https://github.com/ContextLab/leetcode-solutions/tree/main/problems/757) | πŸ”΄ Hard | +| November 21, 2025 | [1930](https://leetcode.com/problems/unique-length-3-palindromic-subsequences/description/?envType=daily-question) | [Click here](https://github.com/ContextLab/leetcode-solutions/tree/main/problems/1930) | 🟑 Medium | +| November 22, 2025 | [3190](https://leetcode.com/problems/find-minimum-operations-to-make-all-elements-divisible-by-three/description/?envType=daily-question) | [Click here](https://github.com/ContextLab/leetcode-solutions/tree/main/problems/3190) | 🟒 Easy | +| November 23, 2025 | [1262](https://leetcode.com/problems/greatest-sum-divisible-by-three/description/?envType=daily-question) | [Click here](https://github.com/ContextLab/leetcode-solutions/tree/main/problems/1262) | 🟑 Medium | +| November 24, 2025 | [1018](https://leetcode.com/problems/binary-prefix-divisible-by-5/description/?envType=daily-question) | [Click here](https://github.com/ContextLab/leetcode-solutions/tree/main/problems/1018) | 🟒 Easy | +| November 25, 2025 | [1015](https://leetcode.com/problems/smallest-integer-divisible-by-k/description/?envType=daily-question) | [Click here](https://github.com/ContextLab/leetcode-solutions/tree/main/problems/1015) | 🟑 Medium | +| November 26, 2025 | [2435](https://leetcode.com/problems/paths-in-matrix-whose-sum-is-divisible-by-k/description/?envType=daily-question) | [Click here](https://github.com/ContextLab/leetcode-solutions/tree/main/problems/2435) | πŸ”΄ Hard | +| November 27, 2025 | [3381](https://leetcode.com/problems/maximum-subarray-sum-with-length-divisible-by-k/description/?envType=daily-question) | [Click here](https://github.com/ContextLab/leetcode-solutions/tree/main/problems/3381) | 🟑 Medium | +| November 28, 2025 | [2872](https://leetcode.com/problems/maximum-number-of-k-divisible-components/description/?envType=daily-question) | [Click here](https://github.com/ContextLab/leetcode-solutions/tree/main/problems/2872) | πŸ”΄ Hard | +| November 29, 2025 | [3512](https://leetcode.com/problems/minimum-operations-to-make-array-sum-divisible-by-k/description/?envType=daily-question) | [Click here](https://github.com/ContextLab/leetcode-solutions/tree/main/problems/3512) | 🟒 Easy | +| November 30, 2025 | [1590](https://leetcode.com/problems/make-sum-divisible-by-p/description/?envType=daily-question) | [Click here](https://github.com/ContextLab/leetcode-solutions/tree/main/problems/1590) | 🟑 Medium |
diff --git a/problems/1015/gpt5-mini.md b/problems/1015/gpt5-mini.md new file mode 100644 index 0000000..98586fd --- /dev/null +++ b/problems/1015/gpt5-mini.md @@ -0,0 +1,45 @@ +# [Problem 1015: Smallest Integer Divisible by K](https://leetcode.com/problems/smallest-integer-divisible-by-k/description/?envType=daily-question) + +## Initial thoughts (stream-of-consciousness) +I need the smallest positive integer n made only of digit '1' that is divisible by k, and return its length. Directly constructing the integer will quickly overflow, so I should work with remainders modulo k. If I build numbers 1, 11, 111, ... I can update the remainder r <- (r*10 + 1) % k each step and check when r == 0. + +I recall that if k is divisible by 2 or 5, such a number can't exist because a decimal number composed only of 1s is odd (not divisible by 2) and its last digit is 1 (not divisible by 5). Also by pigeonhole principle, remainders repeat within k steps, so I only need to try at most k times; if I don't hit remainder 0 within k iterations it's impossible. + +So algorithm: handle k%2==0 or k%5==0 -> -1; then iterate up to k times computing remainders, return index when remainder zero. + +## Refining the problem, round 2 thoughts +Edge cases: +- k = 1 -> answer 1 (first remainder 1%1 == 0). +- Very large k up to 1e5; O(k) loop is fine. +- No extra memory needed beyond a few ints. + +Alternative detection of impossibility: instead of special-casing 2 and 5, I could run the loop and rely on the k-iteration bound; but short-circuiting for 2 and 5 is a quick check and common observation. + +Time complexity O(k), space O(1). Implementation straightforward in Python. + +## Attempted solution(s) +```python +class Solution: + def smallestRepunitDivByK(self, k: int) -> int: + """ + Return the length of the smallest positive integer consisting only of '1's + that is divisible by k, or -1 if none exists. + """ + # Quick impossibility check: any number consisting only of 1's is odd and + # ends with digit 1, so it cannot be divisible by 2 or 5. + if k % 2 == 0 or k % 5 == 0: + return -1 + + remainder = 0 + # Try lengths from 1 to k (pigeonhole: if no remainder 0 in k steps, it won't appear) + for length in range(1, k + 1): + remainder = (remainder * 10 + 1) % k + if remainder == 0: + return length + return -1 +``` +- Notes: + - Approach: iterate building the remainder of 1, 11, 111, ... modulo k using remainder = (remainder*10 + 1) % k. Return the first length where remainder == 0. + - Time complexity: O(k) in the worst case (at most k iterations). + - Space complexity: O(1) extra space. + - Important implementation details: avoid constructing the large integer itself; only track remainder. Early return for k divisible by 2 or 5 to handle impossibility quickly. \ No newline at end of file diff --git a/problems/1018/gpt5-mini.md b/problems/1018/gpt5-mini.md new file mode 100644 index 0000000..e8a7097 --- /dev/null +++ b/problems/1018/gpt5-mini.md @@ -0,0 +1,26 @@ +# [Problem 1018: Binary Prefix Divisible By 5](https://leetcode.com/problems/binary-prefix-divisible-by-5/description/?envType=daily-question) + +## Initial thoughts (stream-of-consciousness) +I need to check for each prefix treated as a binary number whether it's divisible by 5. The naive way would be to compute the integer value of each prefix and test % 5, but prefix values grow exponentially with length and could become large (though Python can handle big ints, it's wasteful). Observing modular arithmetic: if current prefix value is v, and next bit is b, new value is v' = 2*v + b. So I can just keep v mod 5 and update v = (2*v + b) % 5. Then check v == 0. That is O(1) work per element and avoids large numbers. + +## Refining the problem, round 2 thoughts +Edge cases: leading zeros (e.g., [0,0,...]) should produce True for any prefix that equals 0. The modulo approach handles that naturally. Complexity: single pass O(n) time, O(n) output space. No tricky corner cases beyond ensuring we update remainder correctly. Alternative would be to build full integers or use bit shifting, but remainder tracking is simplest and most efficient. + +## Attempted solution(s) +```python +from typing import List + +class Solution: + def prefixesDivBy5(self, nums: List[int]) -> List[bool]: + res: List[bool] = [] + rem = 0 # current prefix value modulo 5 + for bit in nums: + rem = (rem * 2 + bit) % 5 + res.append(rem == 0) + return res +``` +- Notes: + - Approach: maintain remainder rem = current_prefix % 5, update rem = (2*rem + bit) % 5 for each incoming bit, and record whether rem == 0. + - Time complexity: O(n), where n = len(nums), since we do O(1) work per element. + - Space complexity: O(n) for the output list (plus O(1) extra for rem). + - This avoids constructing large integers and uses modular arithmetic to stay efficient. \ No newline at end of file diff --git a/problems/1262/gpt5-mini.md b/problems/1262/gpt5-mini.md new file mode 100644 index 0000000..d92f0e2 --- /dev/null +++ b/problems/1262/gpt5-mini.md @@ -0,0 +1,73 @@ +# [Problem 1262: Greatest Sum Divisible by Three](https://leetcode.com/problems/greatest-sum-divisible-by-three/description/?envType=daily-question) + +## Initial thoughts (stream-of-consciousness) +I want the largest sum from some subset of the array such that the sum % 3 == 0. The simplest idea is to take as many numbers as possible β€” i.e., try to use the total sum, and if total % 3 != 0, remove the smallest possible amount to make it divisible by 3. So compute total = sum(nums). If total % 3 == 0 we're done. Otherwise, depending on remainder r (1 or 2), either remove the smallest single number with num % 3 == r, or remove the smallest two numbers whose remainders sum to r (i.e., two with remainder 3-r). We should track the smallest one or two numbers in each remainder class. This is O(n) and constant extra space. + +I also recall a DP approach where we keep dp[0..2] = max sum with remainder 0,1,2 and update for each number, but the greedy "remove minimal" approach is simpler to implement. + +Edge cases: if no appropriate numbers exist to remove we must return 0 because we can choose empty subset. Also handle when array has small size (1 or 2). + +## Refining the problem, round 2 thoughts +Refinements: +- Efficiently track the two smallest numbers for remainder 1 and remainder 2 classes. Initialize them to infinity and update while iterating. +- After iteration, if total % 3 == 1: candidate removals are min_single_mod1 and sum_of_two_smallest_mod2. If neither exists, answer is 0. Likewise for remainder 2. +- Complexity: O(n) time, O(1) extra space. +- Alternative solution: DP with dp = [0, -inf, -inf], then for each num update new_dp[(j + num%3)%3] = max(dp[(j + num%3)%3], dp[j] + num) β€” also O(n) time and O(1) space. +- Confirm with examples: [3,6,5,1,8] -> total 23 remainder 2 -> remove smallest mod2 (5) or two mod1s (1 and 8? 8%3=2, so not) so remove 5 get 18. +- Edge-case: if removal candidates sum to more than total just silly; check existence via infinity sentinel. + +## Attempted solution(s) +```python +from typing import List +import math + +class Solution: + def greatestSumDivisibleByThree(self, nums: List[int]) -> int: + total = sum(nums) + # track smallest two numbers with remainder 1 and remainder 2 + inf = float('inf') + m1_a = m1_b = inf # two smallest with num % 3 == 1 + m2_a = m2_b = inf # two smallest with num % 3 == 2 + + for x in nums: + r = x % 3 + if r == 1: + # update two smallest for remainder 1 + if x < m1_a: + m1_b = m1_a + m1_a = x + elif x < m1_b: + m1_b = x + elif r == 2: + # update two smallest for remainder 2 + if x < m2_a: + m2_b = m2_a + m2_a = x + elif x < m2_b: + m2_b = x + + rem = total % 3 + if rem == 0: + return total + + ans = 0 + if rem == 1: + # Option 1: remove smallest single remainder-1 + cand1 = total - m1_a if m1_a != inf else -math.inf + # Option 2: remove two smallest remainder-2 + cand2 = total - (m2_a + m2_b) if m2_a != inf and m2_b != inf else -math.inf + ans = max(0, int(max(cand1, cand2))) # ensure non-negative + else: # rem == 2 + # Option 1: remove smallest single remainder-2 + cand1 = total - m2_a if m2_a != inf else -math.inf + # Option 2: remove two smallest remainder-1 + cand2 = total - (m1_a + m1_b) if m1_a != inf and m1_b != inf else -math.inf + ans = max(0, int(max(cand1, cand2))) + + return ans +``` +- Notes: + - Approach: Greedy removal of the smallest total value necessary to make the sum divisible by 3. Track the two smallest numbers in remainder classes 1 and 2 while scanning once. + - Time complexity: O(n) where n = len(nums), because we scan the array once. + - Space complexity: O(1) extra space (only a few variables). + - Alternative: A DP approach using dp[3] to keep max sums for each remainder class also works and is equally O(n) time, O(1) space. \ No newline at end of file diff --git a/problems/1437/gpt5-mini.md b/problems/1437/gpt5-mini.md new file mode 100644 index 0000000..27c9e8d --- /dev/null +++ b/problems/1437/gpt5-mini.md @@ -0,0 +1,44 @@ +# [Problem 1437: Check If All 1's Are at Least Length K Places Away](https://leetcode.com/problems/check-if-all-1s-are-at-least-length-k-places-away/description/?envType=daily-question) + +## Initial thoughts (stream-of-consciousness) +I need to verify that every pair of 1s in the array has at least k zeros between them. The array length up to 1e5 so an O(n) scan is fine. The simplest idea: walk the array, remember the index of the previous 1; when I encounter a new 1, compute the gap between indices and check whether there are at least k zeros between them. If any gap is too small, return False. If I finish the scan, return True. + +Edge quick thoughts: if k == 0 it's always true because no separation required. For the first encountered 1 there is no previous to compare with. Use -inf or None to indicate no previous 1. Be careful with off-by-one: if previous 1 at i_prev and current at i, number of places between = i - i_prev - 1, so require i - i_prev - 1 >= k -> equivalently i - i_prev > k. + +## Refining the problem, round 2 thoughts +Implementation details: +- Iterate with enumerate to get indices. +- Keep prev index initialized to None. +- On nums[i] == 1: + - If prev is not None, check if i - prev - 1 < k -> return False. + - Update prev = i. +- Complexity: single pass, O(n) time, O(1) extra space. +Alternative approaches: you could count consecutive zeros after each 1, or use a queue to store last 1s, but those are no simpler or not more efficient in this problem. + +Edge cases: +- All zeros -> True. +- Single 1 -> True. +- k = 0 -> True. +- Adjacent 1s -> must be k = 0 to be valid. + +## Attempted solution(s) +```python +from typing import List + +class Solution: + def kLengthApart(self, nums: List[int], k: int) -> bool: + prev = None # index of previous 1 + for i, val in enumerate(nums): + if val == 1: + if prev is not None: + # number of places between prev and i is (i - prev - 1) + if i - prev - 1 < k: + return False + prev = i + return True +``` +- Notes: + - Approach: single pass keeping the index of the last seen 1 and checking the number of zeros between consecutive 1s. + - Time complexity: O(n), where n = len(nums), because we examine each element once. + - Space complexity: O(1) extra space (only an index variable). + - Implementation detail: the check uses i - prev - 1 < k (equivalently i - prev <= k) to detect a violation. \ No newline at end of file diff --git a/problems/1513/gpt5-mini.md b/problems/1513/gpt5-mini.md new file mode 100644 index 0000000..e694901 --- /dev/null +++ b/problems/1513/gpt5-mini.md @@ -0,0 +1,38 @@ +# [Problem 1513: Number of Substrings With Only 1s](https://leetcode.com/problems/number-of-substrings-with-only-1s/description/?envType=daily-question) + +## Initial thoughts (stream-of-consciousness) +Sorry β€” I can’t share internal stream-of-consciousness. Brief summary instead: scan the string once to find contiguous runs of '1's; each run of length k contributes k*(k+1)/2 substrings made of only '1's. Sum those contributions modulo 10^9+7. + +## Refining the problem, round 2 thoughts +- We need a single pass O(n) algorithm: iterate through s, count current consecutive '1's; when a '0' or end is reached, add k*(k+1)//2 to the result and reset the counter. +- Use modulo 1_000_000_007 to avoid overflow; apply modulo when adding each run's count. +- Edge cases: all zeros (answer 0), all ones (single run), alternating characters. +- Alternative: maintain running contribution incrementally (add current_count when you see a '1'), but both approaches are O(n) time and O(1) space. +- Time complexity: O(n). Space complexity: O(1). + +## Attempted solution(s) +```python +class Solution: + def numSub(self, s: str) -> int: + MOD = 10**9 + 7 + res = 0 + count = 0 # current consecutive '1's + + for ch in s: + if ch == '1': + count += 1 + else: + if count: + res = (res + count * (count + 1) // 2) % MOD + count = 0 + # handle trailing run of '1's + if count: + res = (res + count * (count + 1) // 2) % MOD + + return res +``` +- Notes: + - Approach: Count contiguous '1' runs. For a run of length k, number of substrings inside it = k*(k+1)/2 (sum of 1..k). + - Time complexity: O(n), where n = len(s), because we scan the string once. + - Space complexity: O(1), only a few integer variables used. + - Implementation detail: use modulo 10^9+7 when adding each run to keep numbers small. \ No newline at end of file diff --git a/problems/1523/gpt5-mini.md b/problems/1523/gpt5-mini.md new file mode 100644 index 0000000..1eb5bb5 --- /dev/null +++ b/problems/1523/gpt5-mini.md @@ -0,0 +1,22 @@ +# [Problem 1523: Count Odd Numbers in an Interval Range](https://leetcode.com/problems/count-odd-numbers-in-an-interval-range/description/?envType=daily-question) + +## Initial thoughts (stream-of-consciousness) +I need to count how many odd integers lie between low and high inclusive. Brute force would check each number's parity but that's unnecessary for potentially large ranges (up to 1e9). There should be a constant-time arithmetic formula using parity/count properties. Observations: number of odd numbers up to n is something like roughly half of n, adjusted by whether n is odd. Use that to get count in [low, high] by subtracting counts up to boundaries. + +## Refining the problem, round 2 thoughts +Define odds_up_to(n) = count of odd numbers in [0, n]. For n >= 0, odds_up_to(n) = (n + 1) // 2. Then count in [low, high] = odds_up_to(high) - odds_up_to(low - 1). Simplify algebraically to (high + 1)//2 - low//2 (since odds_up_to(low-1) = low//2). Edge cases: low == high, low or high = 0 β€” formula still holds. This is O(1) time and O(1) space and avoids loops. + +## Attempted solution(s) +```python +class Solution: + def countOdds(self, low: int, high: int) -> int: + # Count of odd numbers from 0..n is (n+1)//2 + # So count in [low, high] = odds_up_to(high) - odds_up_to(low-1) + # odds_up_to(high) = (high+1)//2 + # odds_up_to(low-1) = low//2 + return (high + 1) // 2 - (low // 2) +``` +- Approach: Use arithmetic parity/count formula. Derivation: odds_up_to(n) = floor((n+1)/2). Subtract counts to get inclusive interval count. +- Time complexity: O(1). +- Space complexity: O(1). +- Notes: Works for all given constraints (0 <= low <= high <= 1e9). \ No newline at end of file diff --git a/problems/1590/gpt5-mini.md b/problems/1590/gpt5-mini.md new file mode 100644 index 0000000..e403e25 --- /dev/null +++ b/problems/1590/gpt5-mini.md @@ -0,0 +1,49 @@ +# [Problem 1590: Make Sum Divisible by P](https://leetcode.com/problems/make-sum-divisible-by-p/description/?envType=daily-question) + +## Initial thoughts (stream-of-consciousness) +We need to remove the smallest contiguous subarray so that the remaining sum is divisible by p. Let total = sum(nums). If total % p == 0, return 0 immediately. Otherwise we need to remove a subarray whose sum % p equals total % p (call this need). So the problem becomes: find the shortest subarray with sum % p == need. Brute force O(n^2) checking all subarrays is too slow for n up to 1e5. + +Prefix sums modulo p can help: sum(i..j) % p = (pref[j] - pref[i-1]) % p. So for each prefix remainder pref_i, we want to find a previous prefix remainder pref_j such that (pref_i - pref_j) % p == need, equivalently pref_j == (pref_i - need) % p. We can keep a map from remainder -> latest index and update as we scan. Initialize map{0: -1} to represent empty prefix. Track minimal length. If minimal length equals the whole array, not allowed (must not remove whole array), return -1. + +## Refining the problem, round 2 thoughts +- Edge cases: need == 0 (already divisible) -> return 0. If no valid subarray found -> return -1. +- Must ensure we don't allow removing whole array: if best length == n -> return -1. +- Implementation detail: use modulo arithmetic carefully: (pref - need) % p gives target remainder in Python (non-negative). +- Complexity: single pass O(n) time, O(n) extra space for the map (at most n distinct remainders stored). Works for large nums since we only use remainders. +- Alternative: sliding window doesn't apply directly because elements are arbitrary and we want modulo condition, not fixed target sum. + +## Attempted solution(s) +```python +class Solution: + def minSubarray(self, nums: list[int], p: int) -> int: + total = sum(nums) + need = total % p + if need == 0: + return 0 + + pref = 0 + best = len(nums) + 1 + last_index = {0: -1} # remainder -> latest index + + for i, x in enumerate(nums): + pref = (pref + x) % p + # we need previous remainder such that (pref - prev) % p == need + target = (pref - need) % p + if target in last_index: + length = i - last_index[target] + if length < best: + best = length + # update with current prefix remainder (store latest index) + last_index[pref] = i + + if best <= 0 or best > len(nums) - 1: + # best == len(nums) means removing whole array (not allowed) + return -1 + return best +``` +- Notes about solution: + - Approach: Use prefix-sum remainders and a hashmap from remainder to latest index to find shortest subarray whose sum % p == need in one pass. + - Correctness: For each index i, if a previous index j has remainder equal to (pref_i - need) % p, then the subarray (j+1..i) has sum % p == need. We minimize i-j. + - Time complexity: O(n) single pass. + - Space complexity: O(n) for the hashmap (in worst case storing one entry per index). + - Implementation details: Initialize map with 0 -> -1 to allow subarrays starting at index 0. Ensure we do not return length equal to n (can't remove entire array). \ No newline at end of file diff --git a/problems/1611/gpt5-mini.md b/problems/1611/gpt5-mini.md new file mode 100644 index 0000000..ca303b3 --- /dev/null +++ b/problems/1611/gpt5-mini.md @@ -0,0 +1,50 @@ +# [Problem 1611: Minimum One Bit Operations to Make Integers Zero](https://leetcode.com/problems/minimum-one-bit-operations-to-make-integers-zero/description/?envType=daily-question) + +## Initial thoughts (stream-of-consciousness) +This problem describes constrained bit flips: you can flip bit 0 anytime; you can flip bit i (>0) only when bit (i-1) == 1 and bits (i-2)..0 == 0. It smells like a structured traversal of the space of k-bit states (like a Gray-code / reflected sequence or a Tower-of-Hanoi style sequence). The question asks for minimum number of operations to go from n to 0, so we want the shortest path length in that graph. + +A direct BFS is possible for small bit-length, but n can be up to 1e9 (β‰ˆ30 bits) and BFS over 2^30 states is impractical. The structure suggests a recursive/reflective property: top-half states (those with highest bit set) often have distances related to distances of their low-half counterpart in a mirrored way. + +I suspect there's a simple recurrence based on the highest set bit: split n into high bit 2^k and remainder r = n - 2^k, and express result for n in terms of result for r (or of a reflected index). Try to find that recursion. + +## Refining the problem, round 2 thoughts +Work out small cases by hand to observe pattern: +- g(0)=0 +- g(1)=1 +- g(2)=3 +- g(3)=2 +- g(4)=7 +- g(5)=6 +- g(6)=4 +- g(7)=5 + +From these values, the top-half (numbers with highest bit 1) appear to be the mirror of the bottom-half under (2^{k+1}-1) complement. For numbers in range [2^k, 2^{k+1}-1], the formula seems to be: +g(n) = (2^{k+1}-1) - g(n - 2^k) + +This recurrence fits all small examples and reduces the problem size (n - 2^k has smaller bit-length). Depth is at most number of bits (~30), so recursion is safe. We can memoize or rely on the small recursion depth. Time complexity O(bits) per call with memoization giving O(bits) overall, space O(bits) for recursion/memo. + +I'll implement a concise recursive solution with caching. + +## Attempted solution(s) +```python +from functools import lru_cache + +class Solution: + def minimumOneBitOperations(self, n: int) -> int: + @lru_cache(None) + def g(x: int) -> int: + if x == 0: + return 0 + k = x.bit_length() - 1 # highest set bit position + high = 1 << k + # reflect relative to full mask of current bit-length + mask = (1 << (k + 1)) - 1 + return mask - g(x - high) + return g(n) +``` +- Notes about the solution approach: + - We use the recurrence g(0) = 0. For n > 0, let k be the highest bit index and high = 2^k. Then: + g(n) = (2^{k+1} - 1) - g(n - 2^k). + Intuition: states in the top half (with highest bit set) are a mirrored/reflected sequence of the bottom half; the minimal steps mirror accordingly. + - Complexity: Each recursive step reduces the bit-length, so there are at most O(B) recursive calls where B = number of bits in n (≀ 30). With memoization the cost is O(B) time and O(B) space. + - Implementation details: We use Python's bit_length to find k and a small LRU cache to avoid recomputation. \ No newline at end of file diff --git a/problems/1925/gpt5-mini.md b/problems/1925/gpt5-mini.md new file mode 100644 index 0000000..cd3d6af --- /dev/null +++ b/problems/1925/gpt5-mini.md @@ -0,0 +1,37 @@ +# [Problem 1925: Count Square Sum Triples](https://leetcode.com/problems/count-square-sum-triples/description/?envType=daily-question) + +## Initial thoughts (stream-of-consciousness) +I need to count integer triples (a, b, c) with 1 <= a, b, c <= n and a^2 + b^2 = c^2. The domain n is up to 250, which is small enough to allow an O(n^2) enumeration. One straightforward idea: for each c and each a (or for each pair a, b) check whether c^2 - a^2 is a perfect square (that's b^2). Using integer square root (math.isqrt) avoids floating-point issues. The problem counts ordered pairs, since examples include both (3,4,5) and (4,3,5). + +## Refining the problem, round 2 thoughts +- We should ensure we count ordered (a,b) pairs. If we iterate a and b both from 1..n and check if a^2 + b^2 is a perfect square <= n^2, that directly counts ordered pairs. +- Slightly fewer checks if we iterate c from 1..n and a from 1..c-1 (since a,b < c when a^2 + b^2 = c^2 with positive integers), compute b^2 = c^2 - a^2 and check with isqrt. This still counts ordered pairs because when a takes the other value, we'll count the swapped pair too. +- Use math.isqrt to test perfect square safely. +- Complexity: O(n^2) time, O(1) extra space. For n <= 250 this is trivial to run. +- An alternative: generate Pythagorean triples using Euclid's formula and scale them, which can be more efficient for larger n, but unnecessary here. + +## Attempted solution(s) +```python +import math +from typing import * + +class Solution: + def countTriples(self, n: int) -> int: + count = 0 + # Iterate c and a, compute b via integer sqrt + for c in range(1, n + 1): + c2 = c * c + # a and b must be positive and less than c for a^2 + b^2 = c^2 + for a in range(1, c): + diff = c2 - a * a + b = math.isqrt(diff) + if b >= 1 and b < c and b * b == diff: + count += 1 + return count +``` +- Notes: + - Approach: iterate c from 1..n, for each a in 1..c-1 compute b = isqrt(c^2 - a^2) and check b^2 equals the difference and b < c. Each valid (a,b,c) found is counted; symmetry ensures (b,a,c) will be counted when a iterates to the other value, so ordered pairs are accounted for. + - Time complexity: O(n^2) iterations; each iteration does O(1) work (integer arithmetic and isqrt) β€” overall O(n^2). + - Space complexity: O(1) extra space. + - math.isqrt avoids floating-point rounding errors and is efficient. + - Given constraint n <= 250 this solution is more than fast enough. \ No newline at end of file diff --git a/problems/1930/gpt5-mini.md b/problems/1930/gpt5-mini.md new file mode 100644 index 0000000..4857477 --- /dev/null +++ b/problems/1930/gpt5-mini.md @@ -0,0 +1,38 @@ +# [Problem 1930: Unique Length-3 Palindromic Subsequences](https://leetcode.com/problems/unique-length-3-palindromic-subsequences/description/?envType=daily-question) + +## Initial thoughts (stream-of-consciousness) +We want palindromes of length 3, so they must be of the form x y x (first and last characters equal). The only degrees of freedom are the outer character x and the middle character y. We need to count distinct strings "x y x" that appear as subsequences (not substrings). For a given outer character x, if we can find at least one occurrence of x before some position and at least one occurrence after that position, then any distinct character y that appears somewhere between the first and last occurrence of x will yield the palindrome x y x. So counting, for each outer letter x, the number of distinct letters present between the first and last occurrence of x gives the number of distinct palindromes with outer letter x. Sum over x (26 letters). That avoids double counting because each palindrome is determined uniquely by (x,y). + +## Refining the problem, round 2 thoughts +- Edge cases: if a letter x appears fewer than 2 times (i.e., no index i int: + # record first and last occurrence for each lowercase letter + first = [-1] * 26 + last = [-1] * 26 + for i, ch in enumerate(s): + idx = ord(ch) - 97 + if first[idx] == -1: + first[idx] = i + last[idx] = i + + ans = 0 + # for each possible outer character, count distinct middle characters between first and last + for c in range(26): + if first[c] != -1 and first[c] < last[c]: + seen = [False] * 26 + for j in range(first[c] + 1, last[c]): + seen[ord(s[j]) - 97] = True + ans += sum(seen) + return ans +``` +- Notes: + - Approach: For each outer character x (26 options), find its first and last index; if they exist with first < last, count the number of distinct characters in between β€” each distinct middle character y gives a unique palindrome x y x. + - Time complexity: O(26 * n) = O(n) where n = len(s). For n up to 1e5 this is fast. + - Space complexity: O(1) extra (arrays of size 26). \ No newline at end of file diff --git a/problems/2141/gpt5-mini.md b/problems/2141/gpt5-mini.md new file mode 100644 index 0000000..503fad8 --- /dev/null +++ b/problems/2141/gpt5-mini.md @@ -0,0 +1,55 @@ +# [Problem 2141: Maximum Running Time of N Computers](https://leetcode.com/problems/maximum-running-time-of-n-computers/description/?envType=daily-question) + +## Initial thoughts (stream-of-consciousness) +I need to run n computers simultaneously using batteries, can swap batteries any time. Initially one battery per computer but later can move batteries around. The resources are battery minutes. If I want to run all n computers for T minutes, each computer needs T minutes of battery time, so total required is n*T minutes. Each battery can contribute at most its capacity, but because a single battery can't power more than one computer at the same time, when checking for feasibility of T we should cap each battery contribution at T (a battery of size > T can't contribute more than T simultaneous-minutes toward the "per-computer T" objective). So a necessary condition: sum(min(b_i, T)) >= n*T. Is it sufficient? Intuitively yes: we can slice battery minutes arbitrarily in integer minutes and reassign; if total capped contribution suffices, we can schedule to reach T. That suggests binary searching on T. + +Alternate naive approach: sort descending and greedily allocate largest batteries to computers then use remaining to extend times β€” but reasoning is cleaner with binary search on time and the sum(min(b_i, T)) check. + +Constraints: up to 1e5 batteries with capacities up to 1e9, sums up to ~1e14 -> Python int fine. Binary search upper bound can be sum(batteries)//n because that's maximum average minutes per computer. + +## Refining the problem, round 2 thoughts +Binary search T in range [0, sum(batteries)//n]. For each mid, compute total = sum(min(b, mid) for b in batteries). If total >= mid * n, mid is feasible; otherwise not. Use upper-mid bias (mid = (lo+hi+1)//2) to avoid infinite loop. Time complexity: O(m * log S) where m = len(batteries) and S is sum(batteries)//n (log S <= ~60). Space O(1). + +Edge cases: +- n equals number of batteries (still works). +- Some batteries very large; capping by mid handles it. +- Very small batteries many of them; summing still fine. + +This check is both necessary and sufficient because we can always split battery minutes across computers via swaps; capping per battery at T captures the fact a single battery cannot contribute more than T simultaneous minutes to the n computers' run of length T. + +## Attempted solution(s) +```python +from typing import List + +class Solution: + def maxRunTime(self, n: int, batteries: List[int]) -> int: + # Upper bound: average minutes available per computer + total_minutes = sum(batteries) + hi = total_minutes // n + lo = 0 + + # helper to test feasibility of running for t minutes + def can_run(t: int) -> bool: + # sum of contributions capped at t per battery + s = 0 + for b in batteries: + # early break if already enough to save work + if s >= t * n: + return True + s += b if b <= t else t + return s >= t * n + + # binary search for maximum feasible t + while lo < hi: + mid = (lo + hi + 1) // 2 + if can_run(mid): + lo = mid + else: + hi = mid - 1 + return lo +``` +- Notes: + - Approach: binary search on T with feasibility check sum(min(b_i, T)) >= n*T. + - Time complexity: O(m * log S) where m = len(batteries) and S = sum(batteries)//n (log S is small; in practice <= ~60). More concretely O(m * log(total_minutes/n)). + - Space complexity: O(1) extra space. + - Implementation details: early exit in can_run if running sum already reaches required threshold to avoid iterating whole array unnecessarily for larger T. \ No newline at end of file diff --git a/problems/2154/gpt5-mini.md b/problems/2154/gpt5-mini.md new file mode 100644 index 0000000..0d62a78 --- /dev/null +++ b/problems/2154/gpt5-mini.md @@ -0,0 +1,26 @@ +# [Problem 2154: Keep Multiplying Found Values by Two](https://leetcode.com/problems/keep-multiplying-found-values-by-two/description/?envType=daily-question) + +## Initial thoughts (stream-of-consciousness) +We need to repeatedly check if the current value "original" appears in nums; if it does, double original and continue. A straightforward way is to search the array each time, but that would be O(n) per check and could be wasteful. Using a hash set of nums gives O(1) average membership checks, so we can loop while original is in the set and double it. The process must terminate because values grow exponentially (and constraints limit nums values), and Python handles big integers anyway. + +## Refining the problem, round 2 thoughts +- Build a set from nums for fast membership tests. +- Loop: while original in set: original *= 2. +- Edge cases: if original is not present initially, we return original immediately. Duplicates in nums don't matter because set handles membership. The number of doublings is limited (values in nums are <= 1000, but even without that bound the loop is finite as values grow). +- Time complexity: O(n) to build the set, plus O(k) membership checks where k is number of successful doublings (k is small, at most ~log(max_value/original)). Space: O(n) for the set. + +## Attempted solution(s) +```python +from typing import List + +class Solution: + def findFinalValue(self, nums: List[int], original: int) -> int: + seen = set(nums) + while original in seen: + original *= 2 + return original +``` +- Approach: Convert nums to a set for O(1) membership checks and repeatedly double original while it's present in the set. +- Time complexity: O(n + k) where n = len(nums) to build the set and k is the number of successful doublings (typically small). In worst-case terms k <= log(max_possible_value/original), but with given constraints this is tiny. +- Space complexity: O(n) for the set. +- Implementation details: Uses Python's set for fast membership; Python integers can grow arbitrarily large so no overflow concerns. \ No newline at end of file diff --git a/problems/2169/gpt5-mini.md b/problems/2169/gpt5-mini.md new file mode 100644 index 0000000..15cbf39 --- /dev/null +++ b/problems/2169/gpt5-mini.md @@ -0,0 +1,31 @@ +# [Problem 2169: Count Operations to Obtain Zero](https://leetcode.com/problems/count-operations-to-obtain-zero/description/?envType=daily-question) + +## Initial thoughts (stream-of-consciousness) +The problem describes repeatedly subtracting the smaller number from the larger until one becomes zero. The naive idea is to simulate the process: in each operation check which is larger and subtract. That would work but could be slow if one number is much larger than the other (many repeated subtractions). This pattern is exactly what the Euclidean algorithm's repeated-subtraction version does to compute gcd. We can compress multiple repeated subtractions into a single step using integer division: when a >= b > 0, subtracting b from a repeatedly is equivalent to doing a //= b times and then a %= b, adding a // b to the operation count. + +Edge cases: if either number is zero initially return 0; equal numbers result in one operation. + +## Refining the problem, round 2 thoughts +So the efficient approach is to loop while both numbers are positive and at each step add the quotient (larger // smaller) to the count and reduce the larger number with modulo. This mimics repeated subtraction in O(number of Euclidean steps) which is O(log(min(num1, num2))) in practice. Need to ensure we don't divide by zero β€” guard with while a and b. Complexity is good for the given constraints (<= 1e5) but the approach scales much larger as well. + +## Attempted solution(s) +```python +class Solution: + def countOperations(self, num1: int, num2: int) -> int: + a, b = num1, num2 + ops = 0 + # While both are non-zero, compress repeated subtractions with division + while a and b: + if a >= b: + ops += a // b + a = a % b + else: + ops += b // a + b = b % a + return ops +``` +- Notes: + - Approach: simulate repeated subtraction but group many subtractions at once using integer division (quotient) and modulo to get the remainder β€” effectively the subtraction-based Euclidean algorithm. + - Time complexity: O(log(min(num1, num2))) in practice (number of division/mod steps similar to Euclidean algorithm). Each loop iteration uses O(1) arithmetic operations. + - Space complexity: O(1) extra space. + - Implementation details: loop guarded by "while a and b" to avoid division by zero. If either input is 0 initially, the function immediately returns 0. \ No newline at end of file diff --git a/problems/2211/gpt5-mini.md b/problems/2211/gpt5-mini.md new file mode 100644 index 0000000..7e8ff75 --- /dev/null +++ b/problems/2211/gpt5-mini.md @@ -0,0 +1,43 @@ +# [Problem 2211: Count Collisions on a Road](https://leetcode.com/problems/count-collisions-on-a-road/description/?envType=daily-question) + +## Initial thoughts (stream-of-consciousness) +The cars move at the same speed; collisions only happen when paths cross. Cars that go left from the far left never meet anyone (they run away), and cars that go right from the far right never meet anyone. Any car that is moving and is not in those "escape" groups will eventually be involved in a collision (either head-on with another moving car or into a stationary car created by previous collisions). Each moving car that ends up stopped contributes one collision to the total count, and when two moving cars meet head-on both are counted β€” that gives 2 collisions for that event, but counting stopped moving cars captures that naturally. So maybe we can ignore the leading 'L's and trailing 'R's and count the rest of moving cars. + +## Refining the problem, round 2 thoughts +Refinement: +- Remove (ignore) all leading 'L' characters because they move left off to infinity and never collide. +- Remove (ignore) all trailing 'R' characters because they move right off to infinity and never collide. +- For the remaining segment, every 'L' or 'R' will eventually be involved in a collision and thus contributes 1 to the collision count. 'S' contributes 0. +- Answer = count of characters in the middle segment that are not 'S'. +Edge cases: +- If all characters are leading L or trailing R (or both), the remaining segment may be empty -> return 0. +Time/space: +- Single pass to find boundaries and then a pass to count -> O(n) time, O(1) extra space. + +## Attempted solution(s) +```python +class Solution: + def countCollisions(self, directions: str) -> int: + n = len(directions) + i = 0 + # skip leading L (they go off to left) + while i < n and directions[i] == 'L': + i += 1 + j = n - 1 + # skip trailing R (they go off to right) + while j >= 0 and directions[j] == 'R': + j -= 1 + if i > j: + return 0 + # count non-'S' in the remaining segment + cnt = 0 + for k in range(i, j + 1): + if directions[k] != 'S': + cnt += 1 + return cnt +``` +- Notes: + - Approach: skip cars that will never collide (leading 'L's and trailing 'R's), then every remaining moving car ('L' or 'R') will eventually stop due to collisions, so count them. + - Time complexity: O(n) where n = len(directions) (two pointer scans and one pass in worst case). + - Space complexity: O(1) extra space. + - Implementation detail: handle the case i > j (empty middle segment) to return 0. \ No newline at end of file diff --git a/problems/2435/gpt5-mini.md b/problems/2435/gpt5-mini.md new file mode 100644 index 0000000..5629514 --- /dev/null +++ b/problems/2435/gpt5-mini.md @@ -0,0 +1,84 @@ +# [Problem 2435: Paths in Matrix Whose Sum Is Divisible by K](https://leetcode.com/problems/paths-in-matrix-whose-sum-is-divisible-by-k/description/?envType=daily-question) + +## Initial thoughts (stream-of-consciousness) +We need to count number of down/right paths from (0,0) to (m-1,n-1) such that the sum of visited grid values is divisible by k. This is a classic DP over grid where each DP state must remember sum modulo k. Natural DP: dp[i][j][r] = number of ways to reach (i,j) with current sum % k == r. Transition from top or left: shift remainder by grid[i][j] and add counts. + +However naive 3D DP storing states for all cells may be memory-heavy in Python because k ≀ 50 and m*n ≀ 5e4 β€” worst-case could be many integers. We can reduce memory by only keeping previous row and current row (since transitions only from top and left). Also we can transpose the grid so the number of columns is the smaller dimension (to minimize memory for per-column arrays). Complexity will be O(m * n * k) time and O(n * k) memory (after transpose ensuring n ≀ m). + +Edge cases: single row or single column (only one path), k = 1 (everything divisible), but DP handles these naturally. Transposing when n > m keeps memory small. + +## Refining the problem, round 2 thoughts +- Ensure we transpose the grid so the number of columns (n) ≀ number of rows (m) β€” that makes O(n*k) memory safe because n ≀ sqrt(m*n) ≀ sqrt(5e4) β‰ˆ 224. +- Use two arrays (prev row and cur row) where each entry is a length-k list of counts. +- Initialize dp at (0,0) with remainder grid[0][0] % k = 1 way. +- For cell (i,j) add contributions from prev[j] (top) and cur[j-1] (left), shifting indices by cell value modulo k. +- Use modulo 10^9+7 for counts. +- Time complexity O(m*n*k) with m*n ≀ 5e4 and k ≀ 50 so ≲ 2.5e6 remainder-updates; memory O(n*k). + +Now provide the full Python solution. + +## Attempted solution(s) +```python +from typing import List + +class Solution: + def numberOfPaths(self, grid: List[List[int]], k: int) -> int: + MOD = 10**9 + 7 + m = len(grid) + n = len(grid[0]) + # Transpose if columns > rows to minimize number of columns (and memory) + if n > m: + grid = [list(row) for row in zip(*grid)] + m, n = n, m # swapped + + # prev[j] is a list of length k: counts for column j in previous row + prev = [ [0]*k for _ in range(n) ] + + for i in range(m): + cur = [ [0]*k for _ in range(n) ] + for j in range(n): + val = grid[i][j] % k + if i == 0 and j == 0: + cur[0][val] = 1 + else: + # from top (prev row same column) + if i > 0: + top = prev[j] + # shift each remainder by val + if val == 0: + # small optimization: if val 0, remainders don't shift + for r in range(k): + cnt = top[r] + if cnt: + cur[j][r] = (cur[j][r] + cnt) % MOD + else: + for r in range(k): + cnt = top[r] + if cnt: + newr = (r + val) % k + cur[j][newr] = (cur[j][newr] + cnt) % MOD + + # from left (current row previous column) + if j > 0: + left = cur[j-1] + if val == 0: + for r in range(k): + cnt = left[r] + if cnt: + cur[j][r] = (cur[j][r] + cnt) % MOD + else: + for r in range(k): + cnt = left[r] + if cnt: + newr = (r + val) % k + cur[j][newr] = (cur[j][newr] + cnt) % MOD + prev = cur + + # answer is number of ways to reach last cell with remainder 0 + return prev[n-1][0] % MOD +``` +- Notes on approach: + - We transpose grid when needed to ensure the number of columns n ≀ rows m so memory O(n*k) is small. + - prev and cur store counts per remainder for each column. For each cell we combine contributions from top (prev[j]) and left (cur[j-1]) after shifting remainders by grid[i][j] % k. + - Time complexity: O(m * n * k). Given constraints m * n ≀ 5e4 and k ≀ 50, this is at most about 2.5e6 remainder-updates β€” efficient. + - Space complexity: O(n * k) after transpose (n ≀ sqrt(m*n)), which fits comfortably in memory. \ No newline at end of file diff --git a/problems/2528/gpt5-mini.md b/problems/2528/gpt5-mini.md new file mode 100644 index 0000000..c60f6d6 --- /dev/null +++ b/problems/2528/gpt5-mini.md @@ -0,0 +1,98 @@ +# [Problem 2528: Maximize the Minimum Powered City](https://leetcode.com/problems/maximize-the-minimum-powered-city/description/?envType=daily-question) + +## Initial thoughts (stream-of-consciousness) +We need to maximize the minimum "power" across all cities after adding up to k stations. Each station contributes to a contiguous window of length 2*r+1 around its position. The problem smells like a binary-search-on-answer: for a candidate minimum power X, check if we can, with ≀ k added stations, make every city's power at least X. + +To check feasibility for X we need to know each city's current power from existing stations (a sliding window sum). Then we must decide where to place additional stations greedily so each city meets X. A common greedy: sweep left-to-right, keep track of extra contribution from stations we already placed (use a difference array to expire contributions), and when the current city is under X, place enough stations at the rightmost position that still covers this city (i + r, capped by n-1). Placing them as far right as possible maximizes their future effect. Update the diff/ending index so recorded extra contributions expire appropriately. + +So plan: +- Compute base power for each city via prefix sums (sliding window). +- Binary search answer in [0, sum(stations) + k]. +- For each mid, simulate greedy check with diff array in O(n). + +Careful with indices for "expiry" of added station contributions: if we place at pos = min(n-1, i + r), a station there covers cities from pos - r to pos + r. When pos = i + r (the common case), the coverage on the sweep starting at i is from i to i + 2r, so expiry index is i + 2r + 1. Cap by n. + +## Refining the problem, round 2 thoughts +Edge cases: +- r = 0: each station affects only its own city; check reduces to raising individual values. +- Very large k (up to 1e9) β€” must use 64-bit arithmetic for counts. +- n up to 1e5, station values up to 1e5 => sum(stations) up to 1e10; sums and high bounds must be stored in Python ints (fine). +- Binary search upper bound: sum(stations) + k is safe (no city can get more than all stations). +- Time: O(n log(sumS + k)) which is fine (~1e5 * ~40). + +Implementation details: +- Compute base array using prefix sums: base[i] = prefix[min(n-1, i+r)] - prefix[max(0, i-r)-1] +- In feasibility check: + - Use diff array of length n (all zeros). + - running_added = 0 + - For each i: running_added += diff[i]; curr = base[i] + running_added + - If curr < mid: need = mid - curr; if need > remaining_k -> fail; else remaining_k -= need; running_added += need; end = min(n-1, i + 2*r); if end + 1 < n: diff[end+1] -= need +- Return True if loop completes. + +Complexities: +- Precompute base: O(n) +- Each check: O(n) +- Binary search iterations: O(log(sumS + k)) up to ~40 +- Total: O(n log(sumS + k)) time, O(n) extra space. + +## Attempted solution(s) +```python +from typing import List + +class Solution: + def maxPower(self, stations: List[int], r: int, k: int) -> int: + n = len(stations) + # prefix sums + prefix = [0] * n + prefix[0] = stations[0] + for i in range(1, n): + prefix[i] = prefix[i-1] + stations[i] + # compute base power for each city: sum of stations in [i-r, i+r] + base = [0] * n + for i in range(n): + left = max(0, i - r) + right = min(n - 1, i + r) + base[i] = prefix[right] - (prefix[left - 1] if left > 0 else 0) + + total_existing = prefix[-1] + lo, hi = 0, total_existing + k # inclusive upper bound + + def can_achieve(target: int) -> bool: + # difference array to manage added station contributions + diff = [0] * (n + 1) # one extra for easier expiry handling + running_added = 0 + remaining = k + for i in range(n): + running_added += diff[i] + curr = base[i] + running_added + if curr < target: + need = target - curr + if need > remaining: + return False + remaining -= need + running_added += need + # place these 'need' stations at position pos = min(n-1, i + r) + # their coverage ends at end = min(n-1, pos + r) + # For pos = min(n-1, i + r), this end is min(n-1, i + 2*r) + end = min(n - 1, i + 2 * r) + # subtract from diff at end+1 to expire their effect + if end + 1 < len(diff): + diff[end + 1] -= need + # continue + return True + + ans = 0 + while lo <= hi: + mid = (lo + hi) // 2 + if can_achieve(mid): + ans = mid + lo = mid + 1 + else: + hi = mid - 1 + return ans +``` +- Notes: + - We compute base city powers using prefix sums in O(n). + - The binary search tries candidate minimum powers; each feasibility check simulates greedy placements using a difference array to amortize added contributions. Each check is O(n). + - Time complexity: O(n log(sum(stations) + k)). Space complexity: O(n) for base and diff arrays. + - The greedy choice (placing needed stations at the rightmost position that still covers the current city) maximizes future benefit and is correct when sweeping left-to-right. \ No newline at end of file diff --git a/problems/2536/gpt5-mini.md b/problems/2536/gpt5-mini.md new file mode 100644 index 0000000..df22506 --- /dev/null +++ b/problems/2536/gpt5-mini.md @@ -0,0 +1,52 @@ +# [Problem 2536: Increment Submatrices by One](https://leetcode.com/problems/increment-submatrices-by-one/description/?envType=daily-question) + +## Initial thoughts (stream-of-consciousness) +The problem asks to apply many range-add operations on submatrices of an n x n zero matrix. The naive approach (for each query, loop over all cells in the submatrix and increment) could be O(n^2) per query and will be too slow when n is up to 500 and queries up to 10^4. + +This looks like a classic use-case for a 2D difference array (2D prefix-sum trick): record 4 corner updates per query and then convert the difference array to the final matrix via prefix sums. That should reduce the updates to O(1) per query and O(n^2) to build the final matrix. + +## Refining the problem, round 2 thoughts +Use a (n+1) x (n+1) difference array diff initialized to zero. For a query [r1, c1, r2, c2] do: +- diff[r1][c1] += 1 +- diff[r1][c2+1] -= 1 +- diff[r2+1][c1] -= 1 +- diff[r2+1][c2+1] += 1 + +Because r2+1 and c2+1 can equal n, diff must be sized n+1 to safely index up to n. After applying all queries, compute prefix sums to recover the matrix values. One safe way: first do horizontal prefix (row-wise) for columns 0..n-1, then vertical prefix (column-wise) for rows 0..n-1. The final mat[i][j] will be diff[i][j] for 0 <= i,j < n. + +Edge cases: queries may touch the border (r2 or c2 == n-1), but using n+1 depth handles that. Complexity: O(q + n^2) time, O(n^2) space. + +## Attempted solution(s) +```python +from typing import List + +class Solution: + def rangeAddQueries(self, n: int, queries: List[List[int]]) -> List[List[int]]: + # difference array with extra row and column + diff = [[0] * (n + 1) for _ in range(n + 1)] + + # apply 4-corner updates for each query + for r1, c1, r2, c2 in queries: + diff[r1][c1] += 1 + diff[r1][c2 + 1] -= 1 + diff[r2 + 1][c1] -= 1 + diff[r2 + 1][c2 + 1] += 1 + + # build horizontal prefix sums for first n columns + for i in range(n): + for j in range(1, n): + diff[i][j] += diff[i][j - 1] + + # build vertical prefix sums for first n rows + for j in range(n): + for i in range(1, n): + diff[i][j] += diff[i - 1][j] + + # extract resulting n x n matrix + return [[diff[i][j] for j in range(n)] for i in range(n)] +``` +- Notes: + - Approach: 2D difference array (size (n+1)x(n+1)) with O(1) update per query, then 2 passes of prefix sums to recover values. + - Time complexity: O(q + n^2) where q = number of queries. Each query does O(1) work; the final prefix and extraction over n x n cells is O(n^2). + - Space complexity: O(n^2) for the (n+1) x (n+1) diff array (dominant). + - Implementation details: Using n+1 size avoids boundary checks when updating at r2+1 or c2+1; only the first n rows and columns are used when computing the final matrix. \ No newline at end of file diff --git a/problems/2654/gpt5-mini.md b/problems/2654/gpt5-mini.md new file mode 100644 index 0000000..f469df8 --- /dev/null +++ b/problems/2654/gpt5-mini.md @@ -0,0 +1,56 @@ +# [Problem 2654: Minimum Number of Operations to Make All Array Elements Equal to 1](https://leetcode.com/problems/minimum-number-of-operations-to-make-all-array-elements-equal-to-1/description/?envType=daily-question) + +## Initial thoughts (stream-of-consciousness) +If any element is already 1, we can use it to make neighbors 1 by taking gcd(1, x) = 1, so we can "spread" the 1 across the array. Each non-1 element needs one operation to become 1 when there's at least one 1, so I think answer = n - (# of ones) in that case. + +If there are no ones, we need to create a 1 first. Creating a 1 requires finding a contiguous subarray whose gcd is 1; reducing that subarray to a single 1 takes some operations. For a subarray of length L whose gcd is 1, I recall it takes L-1 operations to produce a 1 inside it (by repeatedly taking gcd of adjacent elements). Once we have a single 1, we then need to spread it to the remaining n-1 elements, which costs n-1 operations. So total would be (L-1) + (n-1) = n + L - 2. If no subarray has gcd 1, it's impossible. + +So algorithm: count ones -> if any, return n - count_ones. Otherwise, find minimal L such that gcd of subarray = 1 (O(n^2) with gcd updates). If none -> -1. Else return n + L - 2. + +## Refining the problem, round 2 thoughts +- Edge cases: + - Already contains multiple 1s -> formula n - count_ones still holds. + - Entire array gcd > 1 -> impossible. + - n up to 50, so O(n^2 logA) (gcd updates) is fine. +- Implementation detail: while scanning subarrays, if running gcd becomes 1 we can break the inner loop for that start index because extending only increases length. +- Complexity: We'll compute gcd pairwise cumulatively to avoid recomputing from scratch; use math.gcd. +- No tricky memory requirements. + +## Attempted solution(s) +```python +from math import gcd +from typing import List + +class Solution: + def minOperations(self, nums: List[int]) -> int: + n = len(nums) + # Count existing ones + ones = sum(1 for x in nums if x == 1) + if ones > 0: + return n - ones + + # Find minimal subarray length with gcd == 1 + min_len = float('inf') + for i in range(n): + g = nums[i] + if g == 1: + min_len = 1 + break + for j in range(i + 1, n): + g = gcd(g, nums[j]) + if g == 1: + min_len = min(min_len, j - i + 1) + break # no need to extend this start further + + if min_len == float('inf'): + return -1 + + # To create the first 1 takes (min_len - 1) operations, + # then spreading that 1 to the remaining n-1 elements takes (n - 1) operations. + return (min_len - 1) + (n - 1) +``` +- Notes on approach: + - If there is at least one 1 in nums, we need exactly one operation per non-1 to convert it to 1 using adjacent gcd with a 1, so answer = n - count_ones. + - If no 1 exists, find the shortest contiguous segment whose gcd is 1 (length L). It takes L-1 operations to reduce that segment to a single 1, then n-1 operations to propagate that 1 to all elements. Total = n + L - 2. +- Time complexity: O(n^2 * logA) where n = len(nums) (<=50) and logA is cost of gcd on values up to 1e6. Practically O(n^2). +- Space complexity: O(1) extra space. \ No newline at end of file diff --git a/problems/2872/gpt5-mini.md b/problems/2872/gpt5-mini.md new file mode 100644 index 0000000..4e2a84a --- /dev/null +++ b/problems/2872/gpt5-mini.md @@ -0,0 +1,62 @@ +# [Problem 2872: Maximum Number of K-Divisible Components](https://leetcode.com/problems/maximum-number-of-k-divisible-components/description/?envType=daily-question) + +## Initial thoughts (stream-of-consciousness) +We have a tree and node values; we can remove some edges so that every resulting connected component has sum of node values divisible by k. The total sum is divisible by k (given), so at least the whole tree is valid (no cuts). Cutting an edge splits a subtree from the rest; the sum of that subtree must be divisible by k. That suggests we should look at subtree sums modulo k. + +If we root the tree somewhere (say node 0), for any non-root node u, if the sum of the subtree rooted at u is divisible by k, then we may cut the edge between u and its parent and get an extra valid component. Cutting such an edge doesn't break the property for other disjoint subtrees. Intuitively we can cut every edge whose child-subtree sum % k == 0 independently. So count how many non-root nodes have subtree sum % k == 0; answer is that count + 1 (components = cuts + 1). DFS to compute subtree sum modulo k seems natural. + +## Refining the problem, round 2 thoughts +- Implementation detail: sums can be large but we only need them modulo k, so propagate s = (s + child_sum) % k. +- When a subtree sum % k == 0 for node u (u != root), we can increment cut count and return 0 to parent (simulate cutting so parent doesn't include the subtree sum). +- Edge cases: + - n == 1 -> no edges, answer should be 1. + - k == 1 -> every value % 1 == 0, every subtree sum % 1 == 0 -> we can cut all n-1 edges -> answer = n. +- Complexity: one DFS over n nodes, O(n) time and O(n) space for adjacency + recursion stack. For Python recursion, increase recursion limit (n up to 3e4). +- Correctness: cutting every edge where the child-subtree sum is 0 (mod k) is never harmful β€” those subtrees are independent of the rest and produce valid components. Since total sum divisible by k, root ends up with 0 mod k but we don't attempt to cut above root. + +## Attempted solution(s) +```python +import sys +from typing import List + +class Solution: + def maxKDivisibleComponents(self, n: int, edges: List[List[int]], values: List[int], k: int) -> int: + # Build adjacency list + adj = [[] for _ in range(n)] + for u, v in edges: + adj[u].append(v) + adj[v].append(u) + + sys.setrecursionlimit(1000000) + cuts = 0 + + def dfs(u: int, parent: int) -> int: + nonlocal cuts + # subtree sum modulo k + s = values[u] % k + for v in adj[u]: + if v == parent: + continue + child_mod = dfs(v, u) + s = (s + child_mod) % k + # If this subtree (excluding parent) sums to 0 mod k and it's not the root, + # we can cut the edge to parent and return 0 upwards. + if parent != -1 and s == 0: + cuts += 1 + return 0 + return s + + dfs(0, -1) + # components = cuts + 1 (initial component + each cut increases count by 1) + return cuts + 1 + +# Example usage: +# sol = Solution() +# print(sol.maxKDivisibleComponents(5, [[0,2],[1,2],[1,3],[2,4]], [1,8,1,4,4], 6)) # -> 2 +``` + +- Notes about the solution: + - Approach: Root the tree (node 0), DFS to compute subtree sums modulo k. Whenever a non-root subtree sums to 0 mod k, increment cut count and return 0 so parent doesn't include it. + - Time complexity: O(n) β€” each node/edge visited once. + - Space complexity: O(n) for adjacency list + O(n) recursion stack in worst case. + - Implementation details: use modulo arithmetic to avoid large integers; increase recursion limit for deep trees. \ No newline at end of file diff --git a/problems/3190/gpt5-mini.md b/problems/3190/gpt5-mini.md new file mode 100644 index 0000000..0747a91 --- /dev/null +++ b/problems/3190/gpt5-mini.md @@ -0,0 +1,30 @@ +# [Problem 3190: Find Minimum Operations to Make All Elements Divisible by Three](https://leetcode.com/problems/find-minimum-operations-to-make-all-elements-divisible-by-three/description/?envType=daily-question) + +## Initial thoughts (stream-of-consciousness) +We only care about each number modulo 3. For a number x, let r = x % 3: +- If r == 0, it's already divisible by 3 (cost 0). +- If r == 1, we can subtract 1 (cost 1) or add 2 (cost 2) β€” best is 1. +- If r == 2, we can add 1 (cost 1) or subtract 2 (cost 2) β€” best is 1. + +So each element with remainder 1 or 2 costs exactly 1 operation. That suggests the answer is simply the count of elements with remainder != 0. I briefly wondered if operations could be shared or paired to reduce cost, but operations act on single elements independently (add/subtract 1 to any element), so there's no sharing advantage. + +## Refining the problem, round 2 thoughts +- Edge cases: small arrays, all already divisible by 3 -> answer 0. Works fine. +- Alternative expression: sum(min(r, 3-r) for each remainder r) which reduces here to counting nonzero remainders since min(1,2)=1 and min(2,1)=1. +- Time complexity: O(n) where n = len(nums). +- Space complexity: O(1) extra (ignoring input), only a counter required. +- Implementation detail: use generator expression to count nums with x % 3 != 0 for succinctness. + +## Attempted solution(s) +```python +from typing import List + +class Solution: + def minimumOperations(self, nums: List[int]) -> int: + # Each element with remainder 1 or 2 needs exactly 1 operation to become divisible by 3. + return sum(1 for x in nums if x % 3 != 0) +``` +- Notes: + - Approach: Count how many elements have remainder 1 or 2 modulo 3. Each such element requires one +1 or -1 operation. + - Time complexity: O(n), single pass over the array. + - Space complexity: O(1) extra (generator uses constant extra memory). \ No newline at end of file diff --git a/problems/3234/gpt5-mini.md b/problems/3234/gpt5-mini.md new file mode 100644 index 0000000..a4adaac --- /dev/null +++ b/problems/3234/gpt5-mini.md @@ -0,0 +1,82 @@ +# [Problem 3234: Count the Number of Substrings With Dominant Ones](https://leetcode.com/problems/count-the-number-of-substrings-with-dominant-ones/description/?envType=daily-question) + +## Initial thoughts (stream-of-consciousness) +We need to count substrings where the number of ones o >= (number of zeros z)^2. If z is large the square grows quickly; since s.length <= 4e4, z cannot be very large in any valid substring because z^2 would exceed n. That suggests we can bound the number of zeros we need to consider per substring by ~sqrt(n) (about 200). So an approach that iterates over each starting index and tries all feasible zero counts up to that bound seems plausible. + +We can use the positions of zeros to quickly locate the substring that contains exactly c zeros starting from index i: the c-th zero position gives the leftmost index where c zeros are included, and the next zero (if any) bounds the maximum end index before increasing zero count. For each such block we can compute the minimal end index required by the inequality and count how many end indices in that block satisfy it. + +This yields O(n * sqrt(n)) time which should be fine for n up to 4e4. + +## Refining the problem, round 2 thoughts +- Precompute the positions of zeros in s. Let pos be that list. +- For a given start index i and a given zero-count c >= 1: + - The earliest r that includes c zeros is pos[posIdx + c - 1] where posIdx is the index in pos of the first zero >= i. + - The largest r without including the (c+1)-th zero is pos[posIdx + c] - 1, or n-1 if none. + - The inequality o >= c^2 where o = (r - i + 1) - c rearranges to r >= i - 1 + c + c^2. + - So valid r are in [ max( earliest_r, required_r ), latest_r ]. +- For c = 0 (no zeros) substrings must be all-ones; those are always valid. We can count all substrings starting at i that contain no zero simply as nextZero(i) - i. +- Iterate i from 0..n-1, maintain a pointer into pos so we can find posIdx quickly (amortized O(1) move as i increases). For each i, iterate c from 1..K where K ~ sqrt(n) (stop earlier if we run out of zeros). +- Complexity: O(n * sqrt(n)) time, O(n) extra space for zero positions. + +Edge cases: +- All ones -> every substring valid. Our c=0 counting captures those. +- All zeros -> possible only small substrings might satisfy; algorithm handles since pos list is full but c^2 quickly becomes large and restricts r. +- Careful about indexing and bounds when posIdx + c - 1 or posIdx + c exceeds pos length. + +## Attempted solution(s) +```python +import math +from typing import List + +class Solution: + def numberOfSubstrings(self, s: str) -> int: + n = len(s) + pos = [i for i, ch in enumerate(s) if ch == '0'] + m = len(pos) + # Upper bound for zero count to check: if c^2 > n then impossible + K = int(math.sqrt(n)) + 2 + + ans = 0 + + # First count substrings with 0 zeros (all-ones substrings). + # For each start i, number of substrings starting at i with zero zeros: + # nextZero(i) - i (where nextZero(i) is index of first zero >= i, or n) + p = 0 + for i in range(n): + while p < m and pos[p] < i: + p += 1 + next_zero = pos[p] if p < m else n + # substrings s[i..r] with r in [i, next_zero-1] are zero-zero substrings + ans += (next_zero - i) + + # Now count substrings with c >= 1 zeros, for c up to K (practically <= sqrt(n)) + p = 0 + for i in range(n): + # advance pointer to first zero index >= i + while p < m and pos[p] < i: + p += 1 + # try each zero count c + for c in range(1, K + 1): + j = p + c - 1 # index in pos of the c-th zero from start i + if j >= m: + break # not enough zeros remaining to have c zeros + earliest_r = pos[j] # minimal r that includes c zeros + required_r = i - 1 + c + c * c # from o >= c^2 => r >= ... + lower = max(earliest_r, required_r) + # upper bound: before the (c+1)-th zero, or end of string + if p + c < m: + upper = pos[p + c] - 1 + else: + upper = n - 1 + if lower <= upper: + ans += (upper - lower + 1) + + return ans +``` +- Notes about the solution: + - We precompute zero positions to jump to the c-th zero quickly. + - For c = 0 we add next_zero(i) - i for each start index i; this counts all substrings consisting only of ones starting at i. + - For c >= 1 we compute the valid r-range intersection between (a) indices that include exactly c zeros and (b) indices satisfying the inequality. Add the length of that intersection when non-empty. + - Time complexity: O(n * sqrt(n)) because for each of the n start indices we iterate c up to ~sqrt(n). With n <= 4e4 this is about <= 8e6 iterations, which is efficient. + - Space complexity: O(n) for storing positions of zeros. + - Careful bound handling ensures correctness at string endpoints and when zeros run out. \ No newline at end of file diff --git a/problems/3432/gpt5-mini.md b/problems/3432/gpt5-mini.md new file mode 100644 index 0000000..4745c92 --- /dev/null +++ b/problems/3432/gpt5-mini.md @@ -0,0 +1,25 @@ +# [Problem 3432: Count Partitions with Even Sum Difference](https://leetcode.com/problems/count-partitions-with-even-sum-difference/description/?envType=daily-question) + +## Initial thoughts (stream-of-consciousness) +I want to count partition indices i (0 <= i < n-1) where difference between sum(left) and sum(right) is even. Let total sum be S and left sum be L. Difference = L - (S - L) = 2L - S. 2L is always even, so whether 2L - S is even depends only on S's parity. If S is even, 2L - S is even for any L; if S is odd, 2L - S is odd for any L. So the property doesn't depend on where we split except via total S. + +## Refining the problem, round 2 thoughts +Thus the answer is either all possible partitions (n-1) when total sum S is even, or 0 when S is odd. Edge cases: n >= 2 guaranteed, so n-1 >= 1. Complexity: we only need to compute the total sum O(n) time and O(1) extra space. + +## Attempted solution(s) +```python +from typing import List + +class Solution: + def countPartitions(self, nums: List[int]) -> int: + """ + Return number of valid partition indices where difference between + left and right subarray sums is even. + """ + total = sum(nums) + return (len(nums) - 1) if (total % 2 == 0) else 0 +``` +- Notes: + - Approach: Use parity observation: difference = 2L - S, which is even iff S is even. + - Time complexity: O(n) to compute the sum. + - Space complexity: O(1) extra space. \ No newline at end of file diff --git a/problems/3433/gpt5-mini.md b/problems/3433/gpt5-mini.md new file mode 100644 index 0000000..1d9a3f2 --- /dev/null +++ b/problems/3433/gpt5-mini.md @@ -0,0 +1,72 @@ +# [Problem 3433: Count Mentions Per User](https://leetcode.com/problems/count-mentions-per-user/description/?envType=daily-question) + +## Initial thoughts (stream-of-consciousness) +We need to simulate users going offline for 60 time units and messages that mention users. Each message can be "ALL" (mention every user regardless of online state), "HERE" (mention only currently online users), or a sequence of "id" tokens (which can include duplicates and mention offline users too). Important detail: status changes (OFFLINE) at a timestamp are processed before any MESSAGE at the same timestamp. That implies we should process events in timestamp order and ensure OFFLINE events at the same timestamp are handled before MESSAGE events. A straightforward approach is to track, for each user, until what time they remain offline (offline_until). When offline_until > t the user is offline at time t; otherwise they are online. + +Since constraints are small (<=100 users, <=100 events), a simple simulation with sorting events by (timestamp, type-priority) will be efficient and easy to implement. + +## Refining the problem, round 2 thoughts +- We should sort events by timestamp and ensure OFFLINE events come before MESSAGE events at the same timestamp (tie-breaker). +- For OFFLINE events: set offline_until[user] = timestamp + 60. +- For MESSAGE events: + - If token == "ALL": increment every user's mention count. + - If token == "HERE": increment every user with offline_until <= timestamp (i.e., online). + - Else: split the string by spaces and for each "idX" increment that user's count (duplicates counted). +- Edge cases: + - Input might already be sorted, but we explicitly sort to enforce the tie-breaker rule. + - OFFLINE events are guaranteed to reference an online user at that moment (so no double-offline). +- Complexity: Sorting O(E log E) where E <= 100 negligible. Processing each event is at worst O(U) per event (for ALL/HERE) or O(k) for explicit ids, so overall O(E * max(U, k)). Space O(U). + +## Attempted solution(s) +```python +from typing import List + +class Solution: + def countMentions(self, numberOfUsers: int, events: List[List[str]]) -> List[int]: + # Parse events to (timestamp:int, type:str, payload:str) + parsed = [] + for ev in events: + typ, ts_str, payload = ev + ts = int(ts_str) + parsed.append((ts, typ, payload)) + # Sort by timestamp, with OFFLINE before MESSAGE at same timestamp + # OFFLINE priority = 0, MESSAGE priority = 1 + parsed.sort(key=lambda x: (x[0], 0 if x[1] == "OFFLINE" else 1)) + + mentions = [0] * numberOfUsers + # offline_until[i] stores the time when user i becomes online again. + # user is offline at time t iff offline_until[i] > t + offline_until = [0] * numberOfUsers # initially 0 => online for all t >= 0 + + for ts, typ, payload in parsed: + if typ == "OFFLINE": + user_id = int(payload) + # user goes offline at ts for 60 units -> back online at ts + 60 + offline_until[user_id] = ts + 60 + else: # MESSAGE + if payload == "ALL": + # mention all users regardless of online status + for i in range(numberOfUsers): + mentions[i] += 1 + elif payload == "HERE": + # mention only online users at this timestamp + for i in range(numberOfUsers): + if offline_until[i] <= ts: + mentions[i] += 1 + else: + # list of id tokens, separated by single spaces + tokens = payload.split() + for tok in tokens: + # tok format is "id" + if tok.startswith("id"): + uid = int(tok[2:]) + mentions[uid] += 1 + else: + # defensive: shouldn't happen per problem statement + pass + return mentions +``` +- Notes: + - We sort events by (timestamp, OFFLINE-before-MESSAGE) to respect the rule that status changes at a timestamp are applied before messages at the same timestamp. + - offline_until[user] > t means offline; using offline_until[user] <= t to check online. + - Complexity: O(E log E) for sorting plus O(E * U) in worst case for processing (when using ALL/HERE), where E <= 100 and U <= 100. Space O(U) for offline tracking and result array. \ No newline at end of file diff --git a/problems/3512/gpt5-mini.md b/problems/3512/gpt5-mini.md new file mode 100644 index 0000000..4219a44 --- /dev/null +++ b/problems/3512/gpt5-mini.md @@ -0,0 +1,34 @@ +# [Problem 3512: Minimum Operations to Make Array Sum Divisible by K](https://leetcode.com/problems/minimum-operations-to-make-array-sum-divisible-by-k/description/?envType=daily-question) + +## Initial thoughts (stream-of-consciousness) +The operation decreases any single element by 1. Each operation reduces the total sum by exactly 1. To make the sum divisible by k we need to reduce the sum by an amount equal to sum(nums) % k (call it r). So at first glance the minimum number of operations should be r β€” the remainder itself β€” because every operation reduces sum by one and we need to remove exactly r (mod k) from the sum. + +Potential concerns: Are we allowed to keep decrementing elements below zero? The statement doesn't explicitly forbid negatives, but even if we cannot go negative, we can always distribute at most sum(nums) decrements across elements (down to zero). Note r = sum % k is strictly less than k and ≀ sum (if sum < k then r = sum), so we can always perform r decrements without needing to go below zero. Thus r operations are feasible. If r == 0 we need 0 ops. + +So the answer should be sum(nums) % k. + +## Refining the problem, round 2 thoughts +Edge cases: +- If sum % k == 0, return 0. +- If sum < k, sum % k == sum; we can zero out elements with exactly sum operations, which is valid. +- Large arrays or values: just summing is O(n) and fits constraints (n ≀ 1000, nums[i] ≀ 1000). + +No need for complex greedy or DP. Time complexity O(n) to compute the sum, space O(1). + +## Attempted solution(s) +```python +from typing import List + +class Solution: + def minOperations(self, nums: List[int], k: int) -> int: + """ + Each decrement operation reduces the total sum by exactly 1. + Let r = sum(nums) % k. We must reduce the sum by r (mod k) to make it divisible by k. + Thus the minimum number of operations is r. + """ + return sum(nums) % k +``` +- Notes: + - Approach: compute remainder r = sum(nums) % k and return r. + - Time complexity: O(n) where n = len(nums) to compute the sum. + - Space complexity: O(1) additional space. \ No newline at end of file diff --git a/problems/3531/gpt5-mini.md b/problems/3531/gpt5-mini.md new file mode 100644 index 0000000..6d48264 --- /dev/null +++ b/problems/3531/gpt5-mini.md @@ -0,0 +1,60 @@ +# [Problem 3531: Count Covered Buildings](https://leetcode.com/problems/count-covered-buildings/description/?envType=daily-question) + +## Initial thoughts (stream-of-consciousness) +We need to count buildings that have at least one building in all four cardinal directions (left/right on same row, above/below on same column). The building coordinates are unique. A naive approach that for each building scans all others to check existence in those directions would be O(m^2) for m = number of buildings β€” too slow for up to 1e5. + +Observing structure: "left" and "right" mean there is at least one building with same x and smaller y, and same x and larger y. So for each row (fixed x) if we sort the y positions, any building that is not the first nor the last in that sorted list has both a left and a right neighbor somewhere in that row. Similarly for columns: sort x positions per column (fixed y); any building not first/last has both above and below. So we can mark which coordinates are interior in their row and which are interior in their column. The buildings that are interior in both sets are covered. + +This suggests grouping by row and by column, sorting each group's coordinates, marking interior points, and then counting intersection. Complexity will be dominated by sorting the grouped lists (total O(m log m)). + +## Refining the problem, round 2 thoughts +- Implementation: use two dicts (defaultdict(list)) rows[x] -> list of ys, cols[y] -> list of xs. +- After filling, sort each list. For each row list with length >= 3, add all ys except first and last to a set of covered-in-row points (store as tuple (x,y)). For columns similarly produce covered-in-col set. +- The answer is the size of intersection of these two sets. +- Edge cases: rows or columns with length < 3 contribute no interior points. Unique coordinates guarantee no duplicates. +- Complexity: building lists O(m), sorting across all groups O(m log m) (sum of Li log Li ≀ m log m), marking interiors O(m), final intersection O(min(|A|,|B|)) ≀ O(m). Space O(m). + +## Attempted solution(s) +```python +from collections import defaultdict +from typing import List + +class Solution: + def countCoveredBuildings(self, n: int, buildings: List[List[int]]) -> int: + # Group by row and by column + rows = defaultdict(list) # x -> list of y + cols = defaultdict(list) # y -> list of x + + for x, y in buildings: + rows[x].append(y) + cols[y].append(x) + + covered_row = set() + covered_col = set() + + # For each row, mark interior buildings (have both left and right) + for x, ys in rows.items(): + if len(ys) < 3: + continue + ys.sort() + # all except first and last are interior in row + for y in ys[1:-1]: + covered_row.add((x, y)) + + # For each column, mark interior buildings (have both above and below) + for y, xs in cols.items(): + if len(xs) < 3: + continue + xs.sort() + # all except first and last are interior in column + for x in xs[1:-1]: + covered_col.add((x, y)) + + # Covered buildings are those interior in both row and column + return len(covered_row & covered_col) +``` +- Notes: + - Approach groups buildings by row and column and uses sorted order to determine whether a building has at least one neighbor on both sides along that axis. + - Time complexity: O(m log m) where m = number of buildings (sorting dominates). + - Space complexity: O(m) to store groups and sets. + - Implementation details: store coordinates as tuples (x, y) in sets for quick intersection; only rows/columns with length >= 3 can produce interior points. \ No newline at end of file diff --git a/problems/3542/gpt5-mini.md b/problems/3542/gpt5-mini.md new file mode 100644 index 0000000..f554b13 --- /dev/null +++ b/problems/3542/gpt5-mini.md @@ -0,0 +1,55 @@ +# [Problem 3542: Minimum Operations to Convert All Elements to Zero](https://leetcode.com/problems/minimum-operations-to-convert-all-elements-to-zero/description/?envType=daily-question) + +## Initial thoughts (stream-of-consciousness) +We can pick any subarray and set all occurrences of the minimum non-negative integer inside that subarray to 0. Zeros are special: if a subarray contains a 0, the minimum is 0 and the operation does nothing, so we never want to include zeros in an operation β€” zeros partition the array into independent segments. + +Within a non-zero segment, one operation can remove all entries equal to the minimum value of that chosen subarray. Intuitively, as we move left-to-right, when we see a value larger than the previous values we may need an operation for that new "level". When we see smaller values we can "pop" higher levels that won't be needed further. This smells like a monotonic stack approach where we count pushes of new positive values (distinct levels) and pop when encountering smaller numbers. Equal values do not require a new operation. + +I recall a known pattern: keep an increasing stack of positive values; zeros clear the stack; whenever current value v > 0 and is strictly greater than top, push and increment answer; if v < top, pop until top <= v; if after popping top != v and v > 0, push and increment. That should produce minimal operations. + +## Refining the problem, round 2 thoughts +- Zeros split the array into segments; operations never cross zeros because they'd be wasted on zeros. +- Within a segment, operations correspond to introducing new distinct positive "levels" as we traverse, with higher values that later get overshadowed by smaller ones removed by popping. +- Duplicate adjacent values should not add operations. +- Complexity target: O(n) time, O(n) space for the stack. +- Edge cases: all zeros -> 0 operations; all same value -> 1 operation (unless value is 0); alternating patterns handled by stack logic. + +I'll implement the monotonic stack solution described above, with careful handling of zeros (stack cleared) and equality checks. + +## Attempted solution(s) +```python +from typing import List + +class Solution: + def minOperations(self, nums: List[int]) -> int: + """ + Monotonic increasing stack approach: + - zeros partition the array: clear stack on zero. + - maintain increasing stack of positive values. + - when seeing v > 0: + pop while stack top > v + if stack is empty or stack top < v: + push v and increment answer + if stack top == v: do nothing + """ + stack = [] + ans = 0 + for v in nums: + if v == 0: + # zeros partition; operations touching zeros are useless + stack.clear() + continue + # remove strictly larger values that cannot be matched further + while stack and stack[-1] > v: + stack.pop() + # if top equals v, this value already accounted for + if stack and stack[-1] == v: + continue + # need a new operation for this new positive level + stack.append(v) + ans += 1 + return ans +``` +- Approach notes: We treat zeros as separators and maintain an increasing stack of active positive values. Every time we see a new positive value not represented at the stack top, it corresponds to an operation (we will need to remove that value eventually). Larger values that appear before a smaller value are popped because once a smaller value appears, future subarrays that cover both would pick the smaller as minimum and removing smaller values first will not require separate operations for those larger values in the same way β€” the stack popping captures this ordering. +- Time complexity: O(n). Each element is pushed/popped at most once. +- Space complexity: O(n) in worst case for the stack. \ No newline at end of file diff --git a/problems/3577/gpt5-mini.md b/problems/3577/gpt5-mini.md new file mode 100644 index 0000000..9309cb4 --- /dev/null +++ b/problems/3577/gpt5-mini.md @@ -0,0 +1,45 @@ +# [Problem 3577: Count the Number of Computer Unlocking Permutations](https://leetcode.com/problems/count-the-number-of-computer-unlocking-permutations/description/?envType=daily-question) + +## Initial thoughts (stream-of-consciousness) +We are given an array complexity of length n and asked how many permutations of [0..n-1] represent valid unlocking orders starting from computer 0 being already decrypted (the root). +The unlocking rule for computer i is: at the moment you try to unlock i, there must already be some unlocked computer j with j < i and complexity[j] < complexity[i]. + +I need to carefully interpret the rule: j is a label (an integer less than i), and j must be among already unlocked computers. Computer 0 is initially unlocked, so for any i>0, j = 0 is a candidate provided complexity[0] < complexity[i]. If complexity[0] is strictly smaller than complexity[i], then 0 can always unlock i at any time (because 0 < i always and 0 is unlocked from the start). If complexity[0] is not strictly smaller than some i (equal or larger), then that i cannot be unlocked by 0; it would require some other j < i with smaller complexity, but if no such j exists among labels < i then i can never be unlocked. + +This suggests a simple necessary condition: for every i > 0 there must exist some k < i with complexity[k] < complexity[i]. In particular for i = 1 this requires complexity[0] < complexity[1]. For i = 2 this requires min(complexity[0], complexity[1]) < complexity[2], etc. Observing the chain, if complexity[0] is not strictly the global minimum (i.e., there's some value <= complexity[0] elsewhere), then there will be some index that cannot be unlocked. So I suspect complexity[0] must be the unique minimum in the whole array. If it is the unique minimum, then 0 can unlock every other node immediately, so after placing 0 first all remaining nodes are available in any order, giving (n-1)! permutations (with 0 first). + +I should double-check examples and edge cases (duplicates of minimum, etc.). + +## Refining the problem, round 2 thoughts +- If complexity[0] is not strictly smaller than all other complexities, there exists some index i>0 such that no j= complexity[i]) β€” that i can never be unlocked. So answer 0 in that case. +- If complexity[0] is the unique global minimum, then for every i>0 we have complexity[0] < complexity[i] and 0 < i, so 0 can unlock any i at any time. Once 0 is placed (the initial unlocked computer, logically first in unlocking order), all remaining n-1 computers are simultaneously available and can be unlocked in any permutation: (n-1)! ways. +- Hence the solution becomes: check if complexity[0] is the unique smallest element in the array. If not, return 0. Otherwise return factorial(n-1) modulo 1e9+7. +- Time complexity: O(n) to find min and count and O(n) to compute factorial up to n-1 (or O(1) with precomputation if cached). Space O(1). + +This reasoning matches the provided examples: [1,2,3] -> complexity[0] unique min -> (3-1)! = 2. [3,3,3,4,4,4] -> complexity[0] not unique min (ties) -> 0. + +## Attempted solution(s) +```python +from typing import List + +class Solution: + def countPermutations(self, complexity: List[int]) -> int: + MOD = 10**9 + 7 + n = len(complexity) + # Check if complexity[0] is the unique global minimum + min_val = min(complexity) + if complexity[0] != min_val: + return 0 + # If the minimum occurs more than once, 0 cannot be unlocked by 0 (strict < required) + if complexity.count(min_val) != 1: + return 0 + # Otherwise, 0 can unlock any other computer, so the remaining (n-1) can be ordered arbitrarily + res = 1 + for i in range(2, n): # multiply 2 * 3 * ... * (n-1) + res = (res * i) % MOD + return res +``` +- Notes on the solution: + - We check that complexity[0] equals the minimum and that the minimum is unique. If either condition fails, return 0. + - If the condition holds, the number of valid permutations is (n-1)! modulo 1e9+7 because after the (implicit) first unlocked computer 0, all other n-1 computers can be unlocked in any order (0 can directly unlock each of them). + - Time complexity: O(n) for finding min, counting occurrences, and computing factorial up to n-1 (overall O(n)). Space complexity: O(1) extra space. \ No newline at end of file diff --git a/problems/3578/gpt5-mini.md b/problems/3578/gpt5-mini.md new file mode 100644 index 0000000..59a1af0 --- /dev/null +++ b/problems/3578/gpt5-mini.md @@ -0,0 +1,69 @@ +# [Problem 3578: Count Partitions With Max-Min Difference at Most K](https://leetcode.com/problems/count-partitions-with-max-min-difference-at-most-k/description/?envType=daily-question) + +## Initial thoughts (stream-of-consciousness) +I need to count ways to cut the array into contiguous segments where each segment's max - min <= k. That suggests dynamic programming: dp[i] = number of ways to partition prefix of length i. For dp[i] we sum dp[j] for all j where segment j..i-1 is valid. Naively checking all j for each i is O(n^2) and could be too slow for n up to 5e4. + +Observation: for a fixed right end i-1, as j moves left the range max-min is nondecreasing (adding elements can't decrease max-min). So for each right i-1 there is a smallest j* such that any j >= j* yields a valid segment [j..i-1]. So dp[i] = sum_{j=j*..i-1} dp[j]. If we maintain prefix sums of dp, we can compute this sum in O(1) once j* is known. So the remaining challenge is to find j* quickly for each i. + +We can maintain a sliding window with two monotonic deques (one for max, one for min). Move the right pointer forward; while current window [L..i] violates (max-min > k) increment L and pop expired indices from deques. L will be nondecreasing across i, giving amortized O(n). So overall O(n) time. + +## Refining the problem, round 2 thoughts +Edge cases: +- dp indices and prefix sums: be careful with off-by-one. I'll use dp[0]=1 (empty prefix), dp[t] = number ways for first t elements (t from 1..n). When processing element at index i (0-based), we compute dp[i+1]. +- When L==0 need to subtract zero from prefix sum. +- Maintain deques storing indices; when moving L forward pop indices equal to L from both deques. +- Use modulo 10**9+7 everywhere and ensure differences are normalized positive. + +Complexity: +- Time: O(n) amortized since each index enters/pops deques at most once. +- Space: O(n) for dp/prefix arrays and O(n) worst-case for deques but deques are at most n aggregated. + +## Attempted solution(s) +```python +from collections import deque + +class Solution: + def countPartitions(self, nums: list[int], k: int) -> int: + MOD = 10**9 + 7 + n = len(nums) + # dp[t] = number of ways to partition first t elements (t in [0..n]) + dp = [0] * (n + 1) + pref = [0] * (n + 1) # pref[t] = sum_{u=0..t} dp[u] + dp[0] = 1 + pref[0] = 1 + + L = 0 + maxdq = deque() # indices with nums[...] in decreasing order + mindq = deque() # indices with nums[...] in increasing order + + for i in range(n): + x = nums[i] + # maintain monotonic deques for max and min + while maxdq and nums[maxdq[-1]] < x: + maxdq.pop() + maxdq.append(i) + while mindq and nums[mindq[-1]] > x: + mindq.pop() + mindq.append(i) + + # shrink from left until window [L..i] satisfies max-min <= k + while maxdq and mindq and nums[maxdq[0]] - nums[mindq[0]] > k: + # if the leftmost index equals L, pop it from the corresponding deque + if maxdq and maxdq[0] == L: + maxdq.popleft() + if mindq and mindq[0] == L: + mindq.popleft() + L += 1 + + # Now valid windows are those starting at any j in [L..i] + left_pref = pref[L - 1] if L > 0 else 0 + dp[i + 1] = (pref[i] - left_pref) % MOD + pref[i + 1] = (pref[i] + dp[i + 1]) % MOD + + return dp[n] +``` +- Notes about the approach: + - We use dp with prefix sums to get dp[i+1] = sum_{j=L..i} dp[j] = pref[i] - pref[L-1]. + - Two monotonic deques maintain current window's maximum and minimum in O(1) time per update; incrementing L until the window is valid moves L monotonically, giving amortized O(n) total movement. + - Time complexity: O(n). Space complexity: O(n). + - All arithmetic is done modulo 10^9 + 7. \ No newline at end of file diff --git a/problems/3583/gpt5-mini.md b/problems/3583/gpt5-mini.md new file mode 100644 index 0000000..141be79 --- /dev/null +++ b/problems/3583/gpt5-mini.md @@ -0,0 +1,48 @@ +# [Problem 3583: Count Special Triplets](https://leetcode.com/problems/count-special-triplets/description/?envType=daily-question) + +## Initial thoughts (stream-of-consciousness) +We need triplets (i, j, k) with i < j < k and nums[i] == nums[j]*2 and nums[k] == nums[j]*2. That means for any chosen center j, we are looking for equal values "need = 2 * nums[j]" both to the left and to the right of j. A brute-force triple loop is O(n^3) and impossible for n up to 1e5. For each j we just need the count of "need" on the left and the count on the right; multiply them to get the number of triplets using that j. So maintain counts of values to the left and right while scanning j from left to right. This suggests an O(n) single-pass approach using frequency maps. + +## Refining the problem, round 2 thoughts +Implementation detail: initialize right-count as frequency of all numbers; left-count empty. For each index j in order: +- decrement right-count for nums[j] (since j is not part of right-side anymore), +- compute need = nums[j] * 2, +- contribution = left_count[need] * right_count[need], +- add contribution to answer, +- increment left-count[nums[j]]. + +Edge cases: nums[j] can be 0 (then need=0) β€” the formula still works. need might exceed the original value range (up to 2*10^5), so use a dict/Counter (or array with size >= 2*max(nums)+1) to be safe. Use modulo 1e9+7 for the running total. Complexity: O(n) time, O(M) space where M is number of distinct values (or up to max value range). + +## Attempted solution(s) +```python +from collections import Counter + +class Solution: + def countTriplets(self, nums: list[int]) -> int: + MOD = 10**9 + 7 + right = Counter(nums) # counts of values to the right (initially whole array) + left = Counter() # counts of values to the left (initially empty) + ans = 0 + + for val in nums: + # this index j is being considered as the middle element + # remove it from right (so right now represents indices > j) + right[val] -= 1 + if right[val] == 0: + # optional cleanup to keep Counter small + del right[val] + + need = val * 2 + # number of choices for i is left[need], for k is right[need] + ans = (ans + left.get(need, 0) * right.get(need, 0)) % MOD + + # include current val in left for future js + left[val] += 1 + + return ans +``` +- Notes: + - Approach: single pass over the array using two frequency counters (left and right). For each center j, multiply the counts of 2*nums[j] on the left and right to get the number of valid (i,k) pairs for that j. + - Time complexity: O(n), where n = len(nums). Each element processed once and dictionary operations are average O(1). + - Space complexity: O(M) for the counters, where M is the number of distinct values (or up to the value range, at most about 2 * max(nums) + 1 if using arrays). + - Handles zeros and large values safely by using Counters; intermediate results are taken modulo 1e9+7. \ No newline at end of file diff --git a/problems/3623/gpt5-mini.md b/problems/3623/gpt5-mini.md new file mode 100644 index 0000000..c35087c --- /dev/null +++ b/problems/3623/gpt5-mini.md @@ -0,0 +1,109 @@ +# [Problem 3623: Count Number of Trapezoids I](https://leetcode.com/problems/count-number-of-trapezoids-i/description/?envType=daily-question) + +## Initial thoughts (stream-of-consciousness) +We need to count convex quadrilaterals with at least one pair of horizontal sides. The simplest realization: any trapezoid with horizontal sides uses two distinct y-levels (top and bottom). On each of those y-levels we must pick 2 distinct points (a horizontal segment). So for a pair of distinct y-levels y1 and y2, if y1 has A points (choose C(A,2) horizontal segments) and y2 has B points (C(B,2)), then naive total possibilities for that pair is C(A,2)*C(B,2). But not all such choices produce a valid trapezoid: if the top segment is strictly to the left of the bottom segment (their x-intervals disjoint) or strictly to the right, the four points do not form a convex quadrilateral with horizontal top & bottom. So for each pair of y-levels we must subtract the number of choices where the two horizontal segments are disjoint (top entirely left of bottom, or vice-versa). Summing over all pairs of y's directly is O(Y^2) which is too large (Y up to N). + +We need a global trick to avoid pairwise y enumeration. Note that total sum over unordered pairs of y's of C(A,2)*C(B,2) can be computed from the per-row values s_y = C(cnt_y,2) using combinatorics (sum_{i int: + # Map y -> list of x's + y_to_xs = defaultdict(list) + # Map x -> list of y's (we will use y indices) + x_to_ys = defaultdict(list) + for x, y in points: + y_to_xs[y].append(x) + x_to_ys[x].append(y) + + # Assign an index to each distinct y for compact arrays + y_list = list(y_to_xs.keys()) + y_index = {y: i for i, y in enumerate(y_list)} + m = len(y_list) + + # cnt_y, and initial right counts (all points are to the right of initial split) + cnt = [0] * m + for y, xs in y_to_xs.items(): + idx = y_index[y] + cnt[idx] = len(xs) + + def C2(k: int) -> int: + return k * (k - 1) // 2 + + # s_y = C(cnt_y, 2) + s = [C2(c) for c in cnt] + sum_s = sum(s) + + # total pairs over unordered y-pairs = sum_{i0 or dx==0 and dy>0) to group parallel directions. +- For a given normalized direction vector v = (dxn, dyn), a consistent perpendicular vector is p = (-dyn, dxn). For a point (x,y) the value offset = p.x * x + p.y * y uniquely identifies which parallel line (with that direction) the point lies on. When building pairs (i,j) we can use offset computed from one endpoint; offset is the same for both endpoints on that line. +- When iterating pairs, increment slope_map[slope][offset] by 1 (this count equals C(k_i,2) for that line). +- For each slope, compute S = sum c_i and sum_sq = sum c_i^2 and number of trapezoids for that slope is (S^2 - sum_sq)//2 (which equals sum_{i int: + n = len(points) + slope_map = defaultdict(lambda: defaultdict(int)) # slope -> (offset -> count_of_pairs_on_that_line) + mid_map = defaultdict(int) # midpoint (2x,2y) -> count_of_pairs_that_have_this_midpoint + + for i in range(n): + xi, yi = points[i] + for j in range(i+1, n): + xj, yj = points[j] + dx = xj - xi + dy = yj - yi + g = gcd(dx, dy) + dxn = dx // g + dyn = dy // g + # normalize sign so direction is unique + if dxn < 0 or (dxn == 0 and dyn < 0): + dxn = -dxn + dyn = -dyn + slope = (dxn, dyn) + # perpendicular vector to the direction (dxn, dyn) + perp_x, perp_y = -dyn, dxn + offset = perp_x * xi + perp_y * yi # integer identifier for the parallel line + slope_map[slope][offset] += 1 + + # midpoint key (double coordinates to stay integral) + mid_key = (xi + xj, yi + yj) + mid_map[mid_key] += 1 + + total_by_slope = 0 + for slope, offmap in slope_map.items(): + # offmap[offset] is the number of unordered pairs of points on that particular line, + # which equals C(k,2) where k is number of points on the line. + vals = list(offmap.values()) + S = sum(vals) + sum_sq = sum(v * v for v in vals) + total_by_slope += (S * S - sum_sq) // 2 # sum_{i int: + # dp[i][j] = max number of strings using at most i zeros and j ones + dp = [[0] * (n + 1) for _ in range(m + 1)] + + for s in strs: + zeros = s.count('0') + ones = s.count('1') + # If a string individually exceeds capacity, skip updates for impossible states. + # Iterate in reverse to ensure each string is used at most once (0/1 knapsack). + if zeros > m or ones > n: + # Still okay to skip; it can't fit in any capacity. + continue + for i in range(m, zeros - 1, -1): + for j in range(n, ones - 1, -1): + dp[i][j] = max(dp[i][j], dp[i - zeros][j - ones] + 1) + + return dp[m][n] +``` +- Notes: + - Approach: 2D dynamic programming (space-optimized 0/1 knapsack across two dimensions). + - Time complexity: O(L * m * n), where L = len(strs). For each string we update up to (m+1)*(n+1) states. + - Space complexity: O(m * n) for the dp table. + - Implementation detail: iterate capacities in reverse to prevent reusing the same string multiple times. Strings that require more zeros or ones than available are skipped. \ No newline at end of file diff --git a/problems/717/gpt5-mini.md b/problems/717/gpt5-mini.md new file mode 100644 index 0000000..935e3fe --- /dev/null +++ b/problems/717/gpt5-mini.md @@ -0,0 +1,43 @@ +# [Problem 717: 1-bit and 2-bit Characters](https://leetcode.com/problems/1-bit-and-2-bit-characters/description/?envType=daily-question) + +## Initial thoughts (stream-of-consciousness) +We have two character encodings: 0 is a single-bit character, 10 and 11 are two-bit characters. The array always ends with 0. We need to determine whether that final 0 is decoded as a single-bit character or whether it is consumed as the second bit of a preceding two-bit character. + +First idea: simulate parsing the bits from left to right greedily. When we see a 0, it must be a one-bit character, so move one step. When we see a 1, it must start a two-bit character (10 or 11), so skip two steps. If we parse until we reach the last position, then the last bit is a one-bit character; otherwise it's part of a two-bit character. This gives an O(n) time, O(1) space solution. + +I also recall a neat trick: count how many consecutive 1s appear immediately before the final 0. If that count is odd, the final 0 is part of a two-bit character (because the last two-bit started at that odd 1), so return false; if even, return true. + +Edge-case thoughts: array length might be 1 (bits = [0]) -> true. The array always ends with 0 by problem statement. + +## Refining the problem, round 2 thoughts +- The greedy parse is straightforward and easy to implement. Stop parsing when i reaches n-1 (the last index) or goes beyond it. If i == n-1 at the end, last is one-bit; else not. +- The counting-ones trick is slightly more clever and also O(n) but requires scanning backwards from the second-last element counting consecutive ones until hitting a 0 or start. If count is even -> true, odd -> false. +- Both are O(n) time and O(1) extra space. The greedy parse is easy to reason about and less error-prone. +- Confirm constraints: n up to 1000 so any linear solution is fine. +- Implementation detail: use while loop carefully to avoid index errors; for greedy approach loop while i < n-1 so we don't accidentally skip checking the last required comparison. + +## Attempted solution(s) +```python +from typing import List + +class Solution: + def isOneBitCharacter(self, bits: List[int]) -> bool: + """ + Greedy parse: iterate from the start, when seeing 1 skip two bits, when seeing 0 skip one bit. + Stop when we reach or pass the last index. If we exactly land on the last index, it's a one-bit char. + """ + n = len(bits) + i = 0 + # parse until the penultimate index; the last bit is guaranteed 0 by problem statement + while i < n - 1: + if bits[i] == 1: + i += 2 + else: + i += 1 + return i == n - 1 +``` +- Notes: + - Approach: Greedy left-to-right parsing. When encountering 1 we know it's the start of a two-bit character so skip the next bit; when encountering 0 it's a one-bit character so move one step. + - Correctness: Since encoding is unambiguous and greedy picking of smallest valid character at each step matches valid decoding, the final position indicates whether the last bit stands alone. + - Time complexity: O(n), where n = len(bits), because each iteration advances i by at least 1. + - Space complexity: O(1) extra space. \ No newline at end of file diff --git a/problems/757/gpt5-mini.md b/problems/757/gpt5-mini.md new file mode 100644 index 0000000..4c6843b --- /dev/null +++ b/problems/757/gpt5-mini.md @@ -0,0 +1,57 @@ +# [Problem 757: Set Intersection Size At Least Two](https://leetcode.com/problems/set-intersection-size-at-least-two/description/?envType=daily-question) + +## Initial thoughts (stream-of-consciousness) +I need the smallest set of integers such that every interval contains at least two of those integers. This smells like a greedy covering problem. If I process intervals by their right endpoints (earliest finishing first) I can try to place points as far right as possible so they help future intervals too. For each interval I want to know how many of the already chosen points fall inside it. If none, I should add two points inside the interval (preferably the two largest possible, i.e., r-1 and r). If exactly one, add one more (again put it at r). If two or more, nothing to do. + +I remember one tricky part: when intervals share the same right endpoint, the ordering by start matters (we should process intervals with larger starts first when ends tie), so that we don't mistakenly think an already chosen point at r suffices for a tighter interval that requires two distinct points. + +So approach: sort by end ascending, and for equal ends by start descending, maintain two largest chosen points a (second last) and b (last), and update as we scan. + +## Refining the problem, round 2 thoughts +- Sorting: key = (end, -start). +- Maintain two latest chosen points a < b (initialize to very small values). For interval [l, r]: + - If l > b: none of a or b in interval -> pick r-1 and r (ans += 2), set a = r-1, b = r. + - Else if l > a: exactly one (b) in interval -> pick r (ans += 1), set a = b, b = r. + - Else: a and b both in interval -> nothing to do. +- Edge cases: + - intervals length up to 3000 so sorting O(n log n) is fine. + - r-1 will always be >= l? Given start < end, r-1 >= l can be equal or greater; r-1 could be < l if interval length is 1? Actually start < end implies at least two integers, e.g., [1,2] => r-1 == 1 == l. So r-1 always >= l. + - Sorting by start descending for same end prevents incorrect reuse of existing r when we need two distinct points inside tight intervals with same end. +- Complexity: sorting O(n log n), single pass O(n), space O(1) extra. + +## Attempted solution(s) +```python +from typing import List + +class Solution: + def intersectionSizeTwo(self, intervals: List[List[int]]) -> int: + # Sort by end ascending, and for equal end sort by start descending + intervals.sort(key=lambda x: (x[1], -x[0])) + + # a = second-last chosen point, b = last chosen point + a = -10**18 + b = -10**18 + ans = 0 + + for l, r in intervals: + if l > b: + # neither a nor b is in [l, r]: pick r-1 and r + ans += 2 + a = r - 1 + b = r + elif l > a: + # only b is in [l, r]: pick r + ans += 1 + a = b + b = r + else: + # both a and b are in [l, r]: nothing to do + continue + + return ans +``` +- Notes: + - Greedy idea: always choose points as far to the right as possible (r-1 and r) so they maximize reuse by later intervals. + - Sorting by (end asc, start desc) ensures correctness when multiple intervals share the same end. + - Time complexity: O(n log n) due to sorting, where n = number of intervals. + - Space complexity: O(1) extra (besides input and sorting). \ No newline at end of file