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.

 

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'.


Solution :



title-img 


                            Solution in C :

void visit(char **grid, int i, int j, int gridSize, int *gridColSize)
{
    if (i < 0 || j < 0 || i >= gridSize)
        return;
    if (j >= gridColSize[i])
        return;
    if (grid[i][j] == '0')
        return;
    grid[i][j] = '0';
    visit(grid, i-1, j, gridSize, gridColSize);
    visit(grid, i+1, j, gridSize, gridColSize);
    visit(grid, i, j-1, gridSize, gridColSize);
    visit(grid, i, j+1, gridSize, gridColSize);
}

int numIslands(char** grid, int gridSize, int* gridColSize){
    int i, j;
    int res = 0;
    
    for (i = 0; i < gridSize; i++) {
        for (j = 0; j < gridColSize[i]; j++) {
            if (grid[i][j] == '1') {
                visit(grid, i, j, gridSize, gridColSize);
                res++;
            }
        }
    }
    return res;
}
                        


                        Solution in C++ :

int delRow[] = {1, -1, 0, 0};
int delCol[] = {0, 0, -1, 1};

class Solution {
public:
    int numIslands(vector<vector<char>>& grid) {
        int island = 0;
        int n = grid.size();
        int m = grid[0].size();
        vector<vector<int>> vis(n, vector<int>(m, 0));

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (grid[i][j] == '1' && vis[i][j] == 0) {
                    island++;
                    dfs(grid, vis, n, m, i, j);
                }
            }
        }
        return island;
    }

    void dfs(vector<vector<char>>& grid, vector<vector<int>>& vis, int n, int m, int row, int col) {
        if (row < 0 || row >= n || col < 0 || col >= m || vis[row][col] == 1) {
            return;
        }

        vis[row][col] = 1;

        for (int i = 0; i < 4; i++) {
            int nRow = row + delRow[i];
            int nCol = col + delCol[i];

            if (nRow >= 0 && nRow < n && nCol >= 0 && nCol < m && grid[nRow][nCol] == '1' && vis[nRow][nCol] == 0) {
                dfs(grid, vis, n, m, nRow, nCol);
            }
        }
    }
};
                    


                        Solution in Java :

class Solution {
    private final int[] delRow = {1, -1, 0, 0};
    private final int[] delCol = {0, 0, -1, 1};

    public int numIslands(char[][] grid) {
        int island = 0;
        int n = grid.length;
        int m = grid[0].length;
        int[][] vis = new int[n][m];

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (grid[i][j] == '1' && vis[i][j] == 0) {
                    island++;
                    dfs(grid, vis, n, m, i, j);
                }
            }
        }
        return island;
    }

    private void dfs(char[][] grid, int[][] vis, int n, int m, int row, int col) {
        if (row < 0 || row >= n || col < 0 || col >= m || vis[row][col] == 1) {
            return;
        }

        vis[row][col] = 1;

        for (int i = 0; i < 4; i++) {
            int nRow = row + delRow[i];
            int nCol = col + delCol[i];

            if (nRow >= 0 && nRow < n && nCol >= 0 && nCol < m && grid[nRow][nCol] == '1' && vis[nRow][nCol] == 0) {
                dfs(grid, vis, n, m, nRow, nCol);
            }
        }
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def __init__(self):
        self.delRow = [1, -1, 0, 0]
        self.delCol = [0, 0, -1, 1]

    def numIslands(self, grid):
        island = 0
        n = len(grid)
        m = len(grid[0])
        vis = [[0] * m for _ in range(n)]

        for i in range(n):
            for j in range(m):
                if grid[i][j] == '1' and vis[i][j] == 0:
                    island += 1
                    self.dfs(grid, vis, n, m, i, j)
        return island

    def dfs(self, grid, vis, n, m, row, col):
        if row < 0 or row >= n or col < 0 or col >= m or vis[row][col] == 1:
            return

        vis[row][col] = 1

        for i in range(4):
            nRow = row + self.delRow[i]
            nCol = col + self.delCol[i]

            if 0 <= nRow < n and 0 <= nCol < m and grid[nRow][nCol] == '1' and vis[nRow][nCol] == 0:
                self.dfs(grid, vis, n, m, nRow, nCol)

# Example usage
solution = Solution()
grid = [
    ['1', '1', '1', '1', '0'],
    ['1', '1', '0', '1', '0'],
    ['1', '1', '0', '0', '0'],
    ['0', '0', '0', '0', '0']
]
result = solution.numIslands(grid)
print(result)


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