Added tasks 2546-2598

Author: javadev

README.md

@@ -43,10 +43,10 @@ A **word** is a maximal substring consisting of non-space characters only.
 
 ```kotlin
 class Solution {
-    fun lengthOfLastWord(str: String): Int {
+    fun lengthOfLastWord(s: String): Int {
         var len = 0
-        for (i in str.length - 1 downTo 0) {
-            val ch = str[i]
+        for (i in s.length - 1 downTo 0) {
+            val ch = s[i]
             if (ch == ' ' && len > 0) {
                 break
             } else if (ch != ' ') {

src/main/kotlin/g0001_0100/s0058_length_of_last_word/readme.md

@@ -54,11 +54,11 @@ exection -> execution (insert 'u')
 
 ```kotlin
 class Solution {
-    fun minDistance(w1: String, w2: String): Int {
-        val n1 = w1.length
-        val n2 = w2.length
+    fun minDistance(word1: String, word2: String): Int {
+        val n1 = word1.length
+        val n2 = word2.length
         if (n2 > n1) {
-            return minDistance(w2, w1)
+            return minDistance(word2, word1)
         }
         val dp = IntArray(n2 + 1)
         for (j in 0..n2) {
@@ -69,7 +69,7 @@ class Solution {
             dp[0] = i
             for (j in 1..n2) {
                 val tmp = dp[j]
-                dp[j] = if (w1[i - 1] != w2[j - 1]) 1 + Math.min(pre, Math.min(dp[j], dp[j - 1])) else pre
+                dp[j] = if (word1[i - 1] != word2[j - 1]) 1 + Math.min(pre, Math.min(dp[j], dp[j - 1])) else pre
                 pre = tmp
             }
         }

src/main/kotlin/g0001_0100/s0072_edit_distance/readme.md

@@ -30,9 +30,10 @@ The solution set **must not** contain duplicate subsets. Return the solution in
 
 ```kotlin
 class Solution {
-    var allComb: MutableList<List<Int>> = ArrayList()
-    var comb: MutableList<Int> = ArrayList()
-    lateinit var nums: IntArray
+    private var allComb: MutableList<List<Int>> = ArrayList()
+    private var comb: MutableList<Int> = ArrayList()
+    private lateinit var nums: IntArray
+
     fun subsetsWithDup(nums: IntArray): List<List<Int>> {
         nums.sort()
         this.nums = nums

src/main/kotlin/g0001_0100/s0090_subsets_ii/readme.md

@@ -35,7 +35,6 @@ Given two integer arrays `preorder` and `inorder` where `preorder` is the preord
 
 ```kotlin
 import com_github_leetcode.TreeNode
-import java.util.HashMap
 
 /*
  * Example:

src/main/kotlin/g0101_0200/s0105_construct_binary_tree_from_preorder_and_inorder_traversal/readme.md

@@ -51,6 +51,7 @@ import kotlin.collections.ArrayList
  */
 class Solution {
     private val order: MutableList<MutableList<Int>> = ArrayList()
+
     fun levelOrderBottom(root: TreeNode?): List<MutableList<Int>> {
         getOrder(root, 0)
         Collections.reverse(order)

src/main/kotlin/g0101_0200/s0107_binary_tree_level_order_traversal_ii/readme.md

@@ -64,7 +64,7 @@ The format of the output is as follows:
 import java.util.Collections
 import java.util.Stack
 
-internal class Solution {
+class Solution {
     internal inner class Node {
         var mem: MutableMap<List<String>, Int> = HashMap()
         fun update(cur: List<String>, cnt: Int) {

src/main/kotlin/g0701_0800/s0770_basic_calculator_iv/readme.md

@@ -49,7 +49,7 @@ import com_github_leetcode.TreeNode
  *     var right: TreeNode? = null
  * }
  */
-internal class Solution {
+class Solution {
     fun rangeSumBST(root: TreeNode?, low: Int, high: Int): Int {
         var ans = 0
         if (root == null) return 0

src/main/kotlin/g0901_1000/s0938_range_sum_of_bst/readme.md

@@ -0,0 +1,62 @@
+[![](https://img.shields.io/github/stars/javadev/LeetCode-in-Kotlin?label=Stars&style=flat-square)](https://github.com/javadev/LeetCode-in-Kotlin)
+[![](https://img.shields.io/github/forks/javadev/LeetCode-in-Kotlin?label=Fork%20me%20on%20GitHub%20&style=flat-square)](https://github.com/javadev/LeetCode-in-Kotlin/fork)
+
+## 2546\. Apply Bitwise Operations to Make Strings Equal
+
+Medium
+
+You are given two **0-indexed binary** strings `s` and `target` of the same length `n`. You can do the following operation on `s` **any** number of times:
+
+*   Choose two **different** indices `i` and `j` where `0 <= i, j < n`.
+*   Simultaneously, replace `s[i]` with (`s[i]` **OR** `s[j]`) and `s[j]` with (`s[i]` **XOR** `s[j]`).
+
+For example, if `s = "0110"`, you can choose `i = 0` and `j = 2`, then simultaneously replace `s[0]` with (`s[0]` **OR** `s[2]` = `0` **OR** `1` = `1`), and `s[2]` with (`s[0]` **XOR** `s[2]` = `0` **XOR** `1` = `1`), so we will have `s = "1110"`.
+
+Return `true` _if you can make the string_ `s` _equal to_ `target`_, or_ `false` _otherwise_.
+
+**Example 1:**
+
+**Input:** s = "1010", target = "0110"
+
+**Output:** true
+
+**Explanation:** We can do the following operations: 
+
+- Choose i = 2 and j = 0. We have now s = "**<ins>0</ins>**0**<ins>1</ins>**0". 
+
+- Choose i = 2 and j = 1. We have now s = "0**<ins>11</ins>**0". Since we can make s equal to target, we return true.
+
+**Example 2:**
+
+**Input:** s = "11", target = "00"
+
+**Output:** false
+
+**Explanation:** It is not possible to make s equal to target with any number of operations.
+
+**Constraints:**
+
+*   `n == s.length == target.length`
+*   <code>2 <= n <= 10<sup>5</sup></code>
+*   `s` and `target` consist of only the digits `0` and `1`.
+
+## Solution
+
+```kotlin
+class Solution {
+    fun makeStringsEqual(s: String, target: String): Boolean {
+        val strLen = s.length
+        var ans1 = false
+        var ans2 = false
+        for (i in 0 until strLen) {
+            if (s[i] == '1') {
+                ans1 = true
+            }
+            if (target[i] == '1') {
+                ans2 = true
+            }
+        }
+        return ans1 == ans2
+    }
+}
+```

src/main/kotlin/g2501_2600/s2546_apply_bitwise_operations_to_make_strings_equal/readme.md

@@ -0,0 +1,105 @@
+[![](https://img.shields.io/github/stars/javadev/LeetCode-in-Kotlin?label=Stars&style=flat-square)](https://github.com/javadev/LeetCode-in-Kotlin)
+[![](https://img.shields.io/github/forks/javadev/LeetCode-in-Kotlin?label=Fork%20me%20on%20GitHub%20&style=flat-square)](https://github.com/javadev/LeetCode-in-Kotlin/fork)
+
+## 2547\. Minimum Cost to Split an Array
+
+Hard
+
+You are given an integer array `nums` and an integer `k`.
+
+Split the array into some number of non-empty subarrays. The **cost** of a split is the sum of the **importance value** of each subarray in the split.
+
+Let `trimmed(subarray)` be the version of the subarray where all numbers which appear only once are removed.
+
+*   For example, `trimmed([3,1,2,4,3,4]) = [3,4,3,4].`
+
+The **importance value** of a subarray is `k + trimmed(subarray).length`.
+
+*   For example, if a subarray is `[1,2,3,3,3,4,4]`, then trimmed(`[1,2,3,3,3,4,4]) = [3,3,3,4,4].`The importance value of this subarray will be `k + 5`.
+
+Return _the minimum possible cost of a split of_ `nums`.
+
+A **subarray** is a contiguous **non-empty** sequence of elements within an array.
+
+**Example 1:**
+
+**Input:** nums = [1,2,1,2,1,3,3], k = 2
+
+**Output:** 8
+
+**Explanation:** We split nums to have two subarrays: [1,2], [1,2,1,3,3]. '
+
+The importance value of [1,2] is 2 + (0) = 2. 
+
+The importance value of [1,2,1,3,3] is 2 + (2 + 2) = 6. 
+
+The cost of the split is 2 + 6 = 8. It can be shown that this is the minimum possible cost among all the possible splits.
+
+**Example 2:**
+
+**Input:** nums = [1,2,1,2,1], k = 2
+
+**Output:** 6
+
+**Explanation:** We split nums to have two subarrays: [1,2], [1,2,1]. 
+
+The importance value of [1,2] is 2 + (0) = 2. 
+
+The importance value of [1,2,1] is 2 + (2) = 4. 
+
+The cost of the split is 2 + 4 = 6. It can be shown that this is the minimum possible cost among all the possible splits.
+
+**Example 3:**
+
+**Input:** nums = [1,2,1,2,1], k = 5
+
+**Output:** 10
+
+**Explanation:** We split nums to have one subarray: [1,2,1,2,1]. 
+
+The importance value of [1,2,1,2,1] is 5 + (3 + 2) = 10. 
+
+The cost of the split is 10. It can be shown that this is the minimum possible cost among all the possible splits.
+
+**Constraints:**
+
+*   `1 <= nums.length <= 1000`
+*   `0 <= nums[i] < nums.length`
+*   <code>1 <= k <= 10<sup>9</sup></code>
+
+## Solution
+
+```kotlin
+class Solution {
+    fun minCost(nums: IntArray, k: Int): Int {
+        val n = nums.size
+        val dp = IntArray(n)
+        dp.fill(-1)
+        val len = Array(n) { IntArray(n) }
+        for (r in len) r.fill(0)
+        for (i in 0 until n) {
+            val count = IntArray(n)
+            count.fill(0)
+            var c = 0
+            for (j in i until n) {
+                count[nums[j]] += 1
+                if (count[nums[j]] == 2) c += 2 else if (count[nums[j]] > 2) c += 1
+                len[i][j] = c
+            }
+        }
+        return f(0, nums, k, len, dp)
+    }
+
+    private fun f(ind: Int, nums: IntArray, k: Int, len: Array<IntArray>, dp: IntArray): Int {
+        if (ind >= nums.size) return 0
+        if (dp[ind] != -1) return dp[ind]
+        dp[ind] = Int.MAX_VALUE
+        for (i in ind until nums.size) {
+            val current = len[ind][i] + k
+            val next = f(i + 1, nums, k, len, dp)
+            dp[ind] = dp[ind].coerceAtMost(current + next)
+        }
+        return dp[ind]
+    }
+}
+```