Number of Islands

Problem Statement

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.

Leetcode link

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'.

Code

Python Code

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

        def turn_to_dust(i,j):
            if (i<0 or j<0 or i==len(grid) or j==len(grid[0]) or grid[i][j]=="0"):
                return
            grid[i][j]="0"
            turn_to_dust(i,j+1)
            turn_to_dust(i,j-1)
            turn_to_dust(i+1,j)
            turn_to_dust(i-1,j)
            return
        res=0
        for i in range(len(grid)):
            for j in range(len(grid[0])):
                if grid[i][j]=="1":
                    res+=1
                    turn_to_dust(i,j)
        return res

C++

class Solution {
public:

    void turn_to_dust(vector<vector<char>>& grid, int i, int j, int m, int n){
        if (i<0 || j<0 || i==m || j==n || grid[i][j]=='0') return;
        grid[i][j]='0';
        turn_to_dust(grid,i+1,j,m,n);
        turn_to_dust(grid,i,j+1,m,n);
        turn_to_dust(grid,i-1,j,m,n);
        turn_to_dust(grid,i,j-1,m,n);
        return;
    }

    int numIslands(vector<vector<char>>& grid) {
        int m = grid.size();
        int n = grid[0].size();
        int res=0;
        for (int i=0; i<m; i++){
            for (int j=0; j<n; j++){
                if (grid[i][j]=='1'){
                    res++;
                    turn_to_dust(grid,i,j,m,n);
                }
            }
        }
        return res;
    }
};