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200. Number Of Islands

Array Depth-First Search Breadth-First Search Union Find Matrix

Problem - Number Of Islands

Medium

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'.

Solutions

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class Solution:
    def numIslands(self, grid: List[List[str]]) -> int:

        rows, cols = len(grid), len(grid[0])

        def dfs(r: int, c: int) -> None:
            if r < 0 or r >= rows or c < 0 or c >= cols:
                return
            if grid[r][c] == '0':
                return

            grid[r][c] = '0'
            dfs(r + 1, c)
            dfs(r - 1, c)
            dfs(r, c + 1)
            dfs(r, c - 1)

        islands = 0
        for r in range(rows):
            for c in range(cols):
                if grid[r][c] == '1':
                    islands += 1
                    dfs(r, c)
        return islands

Submission Stats:

  • Runtime: 226 ms (94.47%)
  • Memory: 19.9 MB (95.64%)