solves problems in python

* 1472. Design Browser History

* problem 211: Design Add and Search Words Data Structure

* added new lines at end of files

* 208: implement Trie (Prefix Tree)

* 2348:  Number of Zero-Filled Subarrays

* 218: The Skyline Problem

* 2492: Minimum Score of a Path Between Two Cities

* added Links + Time & Space Complexity

* 1319: Number of Operations to Make Network Connected

* 200: Number of Islands

Author: sumanth-botlagunta

python/minimum_score_of_a_path_between_two_cities.py

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+# https://leetcode.com/problems/minimum-score-of-a-path-between-two-cities/
+# T: O(N + M) where N is the number of nodes and M is the number of edges
+# S: O(N + M) where N is the number of nodes and M is the number of edges
+
+from collections import deque
+
+class Solution:
+    def __init__(self):
+        self.min_dist = 1e14
+
+    def minScore(self, n: int, roads) -> int:
+        graph = {}
+        vis = [False]*n
+        for x, y, d in roads:
+            if x in graph:
+                graph[x].append((y, d))
+            else:
+                graph[x] = [(y, d)]
+            if y in graph:
+                graph[y].append((x, d))
+            else:
+                graph[y] = [(x, d)]
+
+        visited = set()
+        queue = deque([1])
+
+        while queue:
+            node = queue.popleft()
+            for adj, score in graph[node]:
+                if adj not in visited:
+                    queue.append(adj)
+                    visited.add(adj)
+                self.min_dist = min(self.min_dist, score)
+        return self.min_dist

python/no_of_operations_to_make_network_connected.py

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+# https://leetcode.com/problems/number-of-operations-to-make-network-connected/description/
+# T: O(M) where M is the number of connections
+# S: O(N) where N is the number of nodes
+
+class Solution:
+    def makeConnected(self, n, connections) -> int:
+        if len(connections) < n-1:
+            return -1
+        if n == 1:
+            return 0
+        graph = {}
+        for a, b in connections:
+            if a in graph:
+                graph[a].append(b)
+            else:
+                graph[a] = [b]
+
+            if b in graph:
+                graph[b].append(a)
+            else:
+                graph[b] = [a]
+
+        visited = [0] * n
+
+        def dfs(node):
+            if visited[node]:
+                return 0
+            visited[node] = 1
+            if node in graph:
+                for num in graph[node]:
+                    dfs(num)
+            return 1
+            
+
+        return sum(dfs(node) for node in range(n)) - 1

python/number_of_islands.py

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+# https://leetcode.com/problems/number-of-islands/description/
+# T: O(MN) where M is the number of rows and N is the number of columns.
+# S: O(MN) where M is the number of rows and N is the number of columns.
+
+def numIslands(self, grid):
+    if not grid:
+        return 0
+        
+    count = 0
+    for i in range(len(grid)):
+        for j in range(len(grid[0])):
+            if grid[i][j] == '1':
+                self.dfs(grid, i, j)
+                count += 1
+    return count
+
+def dfs(self, grid, i, j):
+    if i<0 or j<0 or i>=len(grid) or j>=len(grid[0]) or grid[i][j] != '1':
+        return
+    grid[i][j] = '#'
+    self.dfs(grid, i+1, j)
+    self.dfs(grid, i-1, j)
+    self.dfs(grid, i, j+1)
+    self.dfs(grid, i, j-1)

python/number_of_zero_filled_subarrays.py

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+# https://leetcode.com/problems/number-of-zero-filled-subarrays/
+# T: O(N) where N is the length of nums
+# S: O(1)
+
+class Solution:
+    def zeroFilledSubarray(self, nums) -> int:
+        count = 0
+        i = 0
+        while i < len(nums):
+            if nums[i] == 0:
+                j = i
+                while j < len(nums) and nums[j] == 0:
+                    j += 1
+                count += (j - i) * (j - i + 1) // 2
+                i = j
+            else:
+                i = i+1
+
+        return count

python/the_skyline_problem.py

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+# https://leetcode.com/problems/the-skyline-problem/
+# T: O(NlogN) where N is the number of buildings
+# S: O(N) where N is the number of buildings
+
+import bisect
+
+class Solution:
+    def getSkyline(self, buildings):
+        h = []
+        for b in buildings:
+            h.append((b[0], -b[2]))
+            h.append((b[1], b[2]))
+        
+        h.sort()
+        prev = cur = 0
+        m = [0]
+        res = []
+        
+        for i in h:
+            if i[1] < 0:
+                bisect.insort(m, -i[1])
+            else:
+                m.remove(i[1])
+            cur = max(m)
+            if cur != prev:
+                res.append([i[0], cur])
+                prev = cur
+        
+        return res

python/trie.py

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+# https://leetcode.com/problems/implement-trie-prefix-tree/
+# T: O(N) where N is the length of the word
+# S: O(MN) where M is the number of words, N is the maximum length of the word
+
+class TrieNode:
+    def __init__(self):
+        self.children = {}
+        self.is_word = False
+
+class Trie:
+
+    def __init__(self):
+        self.root = TrieNode()
+
+    def insert(self, word: str) -> None:
+        node = self.root
+        for char in word:
+            if char not in node.children:
+                node.children[char] = TrieNode()
+            node = node.children[char]
+        node.is_word = True
+
+    def search(self, word: str) -> bool:
+        node = self.root
+        for char in word:
+            if char not in node.children:
+                return False
+            node = node.children[char]
+        return node.is_word
+        
+
+    def startsWith(self, prefix: str) -> bool:
+        node = self.root
+        for char in prefix:
+            if char not in node.children:
+                return False
+            node = node.children[char]
+        return True