# Number of Islands LeetCode Solution | Easy Approach Share:

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 IslandsSolutions

Time: O(mn)
Space: O(min(m,n))

### C++

``````class Solution {
public:
int numIslands(vector<vector<char>>& grid) {
const int m = grid.size();
const int n = grid.size();
const vector<int> dirs{0, 1, 0, -1, 0};

auto bfs = [&](int r, int c) {
queue<pair<int, int>> q{{{r, c}}};
grid[r][c] = '2';  // mark '2' as visited
while (!q.empty()) {
const auto [i, j] = q.front();
q.pop();
for (int k = 0; k < 4; ++k) {
const int x = i + dirs[k];
const int y = j + dirs[k + 1];
if (x < 0 || x == m || y < 0 || y == n)
continue;
if (grid[x][y] != '1')
continue;
q.emplace(x, y);
grid[x][y] = '2';  // mark '2' as visited
}
}
};

int ans = 0;

for (int i = 0; i < m; ++i)
for (int j = 0; j < n; ++j)
if (grid[i][j] == '1') {
bfs(i, j);
++ans;
}

return ans;
}
};``````

### Java

`````` class Solution {
public int numIslands(char[][] grid) {
int ans = 0;

for (int i = 0; i < grid.length; ++i)
for (int j = 0; j < grid.length; ++j)
if (grid[i][j] == '1') {
bfs(grid, i, j);
++ans;
}

return ans;
}

private static final int[] dirs = {0, 1, 0, -1, 0};

private void bfs(char[][] grid, int r, int c) {
Queue<int[]> q = new ArrayDeque<>();
q.offer(new int[] {r, c});
grid[r][c] = '2'; // mark '2' as visited
while (!q.isEmpty()) {
final int i = q.peek();
final int j = q.poll();
for (int k = 0; k < 4; ++k) {
final int x = i + dirs[k];
final int y = j + dirs[k + 1];
if (x < 0 || x == grid.length || y < 0 || y == grid.length)
continue;
if (grid[x][y] != '1')
continue;
q.offer(new int[] {x, y});
grid[x][y] = '2'; // mark '2' as visited
}
}
}
}
``````

### Python

``````
class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
m = len(grid)
n = len(grid)
dirs = [0, 1, 0, -1, 0]

def bfs(r, c):
q = deque([(r, c)])
grid[r][c] = '2'  # mark '2' as visited
while q:
i, j = q.popleft()
for k in range(4):
x = i + dirs[k]
y = j + dirs[k + 1]
if x < 0 or x == m or y < 0 or y == n:
continue
if grid[x][y] != '1':
continue
q.append((x, y))
grid[x][y] = '2'  # mark '2' as visited

ans = 0

for i in range(m):
for j in range(n):
if grid[i][j] == '1':
bfs(i, j)
ans += 1

return ans``````

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