Number of Islands LeetCode Solution – Queslers

Problem – Number of Islands

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'.

Number of Islands LeetCode Solution in Python

def numIslands(self, grid):
    def sink(i, j):
        if 0 <= i < len(grid) and 0 <= j < len(grid[i]) and grid[i][j] == '1':
            grid[i][j] = '0'
            map(sink, (i+1, i-1, i, i), (j, j, j+1, j-1))
            return 1
        return 0
    return sum(sink(i, j) for i in range(len(grid)) for j in range(len(grid[i])))

Number of Islands LeetCode Solution in Java

public class Solution {
    char[][] g;
    public int numIslands(char[][] grid) {
        int islands = 0;
        g = grid;
        for (int i=0; i<g.length; i++)
            for (int j=0; j<g[i].length; j++)
                islands += sink(i, j);
        return islands;
    }
    int sink(int i, int j) {
        if (i < 0 || i == g.length || j < 0 || j == g[i].length || g[i][j] == '0')
            return 0;
        g[i][j] = '0';
        sink(i+1, j); sink(i-1, j); sink(i, j+1); sink(i, j-1);
        return 1;
    }
}

Number of Islands LeetCode Solution in C++

class Solution {
public:
    int numIslands(vector<vector<char>>& grid) {
        int m = grid.size(), n = m ? grid[0].size() : 0, islands = 0, offsets[] = {0, 1, 0, -1, 0};
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == '1') {
                    islands++;
                    grid[i][j] = '0';
                    queue<pair<int, int>> todo;
                    todo.push({i, j});
                    while (!todo.empty()) {
                        pair<int, int> p = todo.front();
                        todo.pop();
                        for (int k = 0; k < 4; k++) {
                            int r = p.first + offsets[k], c = p.second + offsets[k + 1];
                            if (r >= 0 && r < m && c >= 0 && c < n && grid[r][c] == '1') {
                                grid[r][c] = '0';
                                todo.push({r, c});
                            }
                        }
                    }
                }
            }
        }
        return islands;
    }
};
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