# Count Sub Islands LeetCode Solution

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## Count Sub Islands

You are given two `m x n` binary matrices `grid1` and `grid2` containing only `0`‘s (representing water) and `1`‘s (representing land). An island is a group of `1`‘s connected 4-directionally (horizontal or vertical). Any cells outside of the grid are considered water cells.

An island in `grid2` is considered a sub-island if there is an island in `grid1` that contains all the cells that make up this island in `grid2`.

Return the number of islands in `grid2` that are considered sub-islands.

Example 1:

```Input: grid1 = [[1,1,1,0,0],[0,1,1,1,1],[0,0,0,0,0],[1,0,0,0,0],[1,1,0,1,1]], grid2 = [[1,1,1,0,0],[0,0,1,1,1],[0,1,0,0,0],[1,0,1,1,0],[0,1,0,1,0]]
Output: 3
Explanation: In the picture above, the grid on the left is grid1 and the grid on the right is grid2.
The 1s colored red in grid2 are those considered to be part of a sub-island. There are three sub-islands.
```

Example 2:

```Input: grid1 = [[1,0,1,0,1],[1,1,1,1,1],[0,0,0,0,0],[1,1,1,1,1],[1,0,1,0,1]], grid2 = [[0,0,0,0,0],[1,1,1,1,1],[0,1,0,1,0],[0,1,0,1,0],[1,0,0,0,1]]
Output: 2
Explanation: In the picture above, the grid on the left is grid1 and the grid on the right is grid2.
The 1s colored red in grid2 are those considered to be part of a sub-island. There are two sub-islands.
```

Constraints:

• `m == grid1.length == grid2.length`
• `n == grid1[i].length == grid2[i].length`
• `1 <= m, n <= 500`
• `grid1[i][j]` and `grid2[i][j]` are either `0` or `1`.

Program:

``````int solve(vector<vector<int>>& g1, vector<vector<int>>& g2, int i, int j)
{
int n = g2.size(), m = g2.size();

if(i < 0 || i == n || j <0 || j== m || g2[i][j] == 0)
return 0;

if(g1[i][j] == 0)
return -1;

g2[i][j] = 0;

int a = solve(g1, g2, i+1, j);
int b = solve(g1, g2, i-1, j);
int c = solve(g1, g2, i, j-1);
int d = solve(g1, g2, i, j+1);

if(a == -1 || b == -1 || c == -1 || d == -1)
return -1;

return 1;

}
int countSubIslands(vector<vector<int>>& g1, vector<vector<int>>& g2) {
int n = g2.size(), m = g2.size(), ans =0;

for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
if(g2[i][j] == 1)  {
if(solve(g1, g2, i, j) == 1)
ans++;
}
}
}
return ans;
}``````