dfs-bfs - change the input 😄

Leet_Code/medium/200_Number_of_Islands.py

@@ -0,0 +1,86 @@
+"""
+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'.
+
+Takeaway:
+
+    To solve this question, think like a kindergardener.
+
+    How can you find a single island?
+
+    You look.
+
+    For each 1, you need to check neighbours, this is clearly bfs or dfs.
+
+    Setting the cell to 0 in order to not visit it again is pretty cool.
+"""
+
+class Solution:
+
+    def numIslands(self, grid: list[list[str]]) -> int:
+        m = len(grid)
+        n = len(grid[0])
+        result = 0
+
+        def dfs(i, j):
+            if (i < 0 or 
+                j < 0 or 
+                i >= m or 
+                j >= n or 
+                grid[i][j] == '0'):
+                    return
+			# make the current tile 0
+			# so you will not count this tile again
+            grid[i][j] = '0'
+
+			# go to every direction possible
+            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':
+                    result += 1
+                    dfs(i, j)
+
+        return result