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.

Test Cases

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:

Solution

class Solution {
public:
    int numIslands(vector<vector<char>>& grid) {
        int count = 0, m = grid.size(), n = grid[0].size();
        for(int i=0; i<m; i++) {
            for(int j=0; j<n; j++) {
                if (grid[i][j] == '1') {
                    dfs(grid, i, j, m, n);
                    count++;
                }
            }
        }
        return count;
    }

private:
    void dfs(vector<vector<char>> &grid, int row, int col, int rowsize, int colsize) {
        if (row < 0 || row >= rowsize || col <0 || col >= colsize || grid[row][col] == '0') {
            return;
        }
        grid[row][col] = '0';
        dfs(grid, row-1, col, rowsize, colsize);
        dfs(grid, row+1, col, rowsize, colsize);
        dfs(grid, row, col-1, rowsize, colsize);
        dfs(grid, row, col+1, rowsize, colsize);
    }
};
class Solution {
    public int numIslands(char[][] grid) {
        int count = 0;
        int n = grid.length, m = grid[0].length;
        for(int i=0; i<n; i++) {
            for(int j=0; j<m; j++) {
                if (grid[i][j] == '1') {
                    count++;
                    dfs(grid, i, j, n, m);
                }
            }
        }
        return count;
    }
    
    private void dfs(char[][] grid, int r, int c, int n, int m) {
        grid[r][c] = '2';
        if (r>0 && grid[r-1][c] == '1') dfs(grid, r-1, c, n, m);
        if (r<n-1 && grid[r+1][c] == '1') dfs(grid, r+1, c, n, m);
        if (c>0 && grid[r][c-1] == '1') dfs(grid, r, c-1, n, m);
        if (c<m-1 && grid[r][c+1] == '1') dfs(grid, r, c+1, n, m);
    }
}
class Solution:
    def numIslands(self, grid: List[List[str]]) -> int:
        def graph_search(grid: List[List[str]], i: int, j: int, m: int, n: int):
            if 0 <= i < m and 0 <= j < n and grid[i][j] == '1':
                grid[i][j] = '0'
                graph_search(grid, i+1, j, m, n)
                graph_search(grid, i-1, j, m, n)
                graph_search(grid, i, j+1, m, n)
                graph_search(grid, i, j-1, m, n)
        m: int = len(grid)
        n: int = len(grid[0])
        count = 0
        for i in range(m):
            for j in range(n):
                if grid[i][j] == '1':
                    count += 1
                    graph_search(grid, i, j, m, n)
        return count
Time Complexity: O(mn)
Space Complexity: O(1)