[bhyun-kim, 김병현] Week 9 Solutions

Author: sbk1014

number-of-islands/bhyun-kim.py

@@ -0,0 +1,54 @@
+"""
+200. Number of Islands
+https://leetcode.com/problems/number-of-islands/
+
+Solution:
+    To solve this problem, we can use a depth-first search (DFS) to explore all connected land cells.
+    We can create a helper function that takes the current cell and marks it as visited.
+    
+    - We can iterate through all cells in the grid.
+    - If the current cell is land, we explore all connected land cells using DFS.
+    - We mark all connected land cells as visited.
+    - We increment the island count by 1.
+
+Time complexity: O(m x n)
+    - m and n are the dimensions of the grid.
+    - We explore all cells in the grid once.
+
+Space complexity: O(m x n)
+    - We use a recursive call stack to explore all connected land cells.
+    - The maximum depth of the call stack is the number of cells in the grid.
+
+"""
+
+from typing import List
+
+class Solution:
+    def numIslands(self, grid: List[List[str]]) -> int:
+        if not grid:
+            return 0
+
+        m, n = len(grid), len(grid[0])
+
+        island_count = 0
+
+        def dfs(i, j):
+            if i < 0 or i >= m or j < 0 or j >= n or grid[i][j] != '1':
+                return
+            grid[i][j] = '#'
+
+            dfs(i - 1, j)  
+            dfs(i + 1, j)  
+            dfs(i, j - 1)  
+            dfs(i, j + 1)  
+
+        for i in range(m):
+            for j in range(n):
+
+                if grid[i][j] == '1':
+                    dfs(i, j)
+                    island_count += 1
+
+        return island_count
+
+