Problem #200: Number of Islands

Intuition & Strategy Breakdown

Traverse 2D grid. When encountering '1', increment island count and run BFS/DFS to sink all connected land cells '1' to '0'.

Python Solution

class Solution:
    def numIslands(self, grid: List[List[str]]) -> int:
        if not grid: return 0
        rows, cols = len(grid), len(grid[0])
        count = 0
        def dfs(r, c):
            if r < 0 or r >= rows or c < 0 or c >= cols or 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)
        for r in range(rows):
            for c in range(cols):
                if grid[r][c] == '1':
                    count += 1
                    dfs(r, c)
        return count

Java Solution

class Solution {
    public int numIslands(char[][] grid) {
        if (grid == null || grid.length == 0) return 0;
        int count = 0;
        for (int r = 0; r < grid.length; r++) {
            for (int c = 0; c < grid[0].length; c++) {
                if (grid[r][c] == '1') {
                    count++;
                    dfs(grid, r, c);
                }
            }
        }
        return count;
    }
    private void dfs(char[][] grid, int r, int c) {
        if (r < 0 || r >= grid.length || c < 0 || c >= grid[0].length || grid[r][c] == '0') return;
        grid[r][c] = '0';
        dfs(grid, r + 1, c);
        dfs(grid, r - 1, c);
        dfs(grid, r, c + 1);
        dfs(grid, r, c - 1);
    }
}