number-of-islands sol (py)

Author: hi-rachel

number-of-islands/hi-rachel.py

@@ -0,0 +1,72 @@
+"""
+https://leetcode.com/problems/number-of-islands/
+
+Given an m x n 2D binary grid grid which represents a map of '1's (land) and '0's (water), return the number of islands.
+
+An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.
+
+
+Example 1:
+Input: grid = [
+  ["1","1","1","1","0"],
+  ["1","1","0","1","0"],
+  ["1","1","0","0","0"],
+  ["0","0","0","0","0"]
+]
+Output: 1
+
+Example 2:
+Input: grid = [
+  ["1","1","0","0","0"],
+  ["1","1","0","0","0"],
+  ["0","0","1","0","0"],
+  ["0","0","0","1","1"]
+]
+Output: 3
+
+Constraints:
+m == grid.length
+n == grid[i].length
+1 <= m, n <= 300
+grid[i][j] is '0' or '1'.
+
+BFS 풀이
+TC: O(m * n)
+SC: O(m * n)
+"""
+
+from collections import deque
+from typing import List
+
+class Solution:
+    def numIslands(self, grid: List[List[str]]) -> int:
+        ans = 0
+        dx = [-1, 1, 0, 0]
+        dy = [0, 0, -1, 1]
+        rows = len(grid)
+        cols = len(grid[0])
+
+        def bfs(x, y):
+            queue = deque()
+            queue.append((x, y))
+            grid[x][y] = 2
+
+            while queue:
+                x, y = queue.popleft()
+                for i in range(4):
+                    nx = x + dx[i]
+                    ny = y + dy[i]
+                    if 0 <= nx < rows and 0 <= ny < cols:
+                        if grid[nx][ny] == "1":
+                            grid[nx][ny] = "2"
+                            queue.append((nx, ny))
+
+            return True
+
+        for i in range(rows):
+            for j in range(cols):
+                if grid[i][j] == "1":
+                    if bfs(i, j):
+                        ans += 1
+
+        return ans