# Number of Islands LeetCode Solution

Here, We see Number of Islands LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

# List of all LeetCode Solution

## 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

## Number of Islands LeetCode Solution 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;
}
};```Code language: PHP (php)```

## Number of Islands LeetCode Solution Java

``````public class Solution {
private int n;
private int m;
public int numIslands(char[][] grid) {
int count = 0;
n = grid.length;
if (n == 0) return 0;
m = grid[0].length;
for (int i = 0; i < n; i++){
for (int j = 0; j < m; j++)
if (grid[i][j] == '1') {
DFSMarking(grid, i, j);
++count;
}
}
return count;
}

private void DFSMarking(char[][] grid, int i, int j) {
if (i < 0 || j < 0 || i >= n || j >= m || grid[i][j] != '1') return;
grid[i][j] = '0';
DFSMarking(grid, i + 1, j);
DFSMarking(grid, i - 1, j);
DFSMarking(grid, i, j + 1);
DFSMarking(grid, i, j - 1);
}
}```Code language: PHP (php)```

## Number of Islands LeetCode SolutionJavaScript

``````var numIslands = function(grid) {
let count = 0;
function depthSearch(x, y) {
if (grid[x][y] === '1') {
grid[x][y] = '0';
} else {
return;
}
if (x < grid.length - 1) {
depthSearch(x+1, y);
}
if (y < grid[x].length - 1) {
depthSearch(x, y+1);
}
if (x > 0 && x < grid.length) {
depthSearch(x-1, y);
}
if (y > 0 && y < grid[x].length) {
depthSearch(x, y-1);
}
}
for (let i = 0; i < grid.length; i++) {
for (let j = 0; j < grid[i].length; j++) {
if (grid[i][j] === '1') {
count++;
depthSearch(i, j);
}
}
}
return count;
};```Code language: JavaScript (javascript)```

## Number of Islands LeetCode SolutionPython

``````class Solution(object):
def numIslands(self, grid):
if not grid:
return 0
count = 0
for i in range(len(grid)):
for j in range(len(grid[0])):
if grid[i][j] == '1':
self.dfs(grid, i, j)
count += 1
return count
def dfs(self, grid, i, j):
if i<0 or j<0 or i>=len(grid) or j>=len(grid[0]) or grid[i][j] != '1':
return
grid[i][j] = '#'
self.dfs(grid, i+1, j)
self.dfs(grid, i-1, j)
self.dfs(grid, i, j+1)
self.dfs(grid, i, j-1)``````
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