Count number of islands where every island is row-wise and column-wise separated

Given a rectangular matrix which has only two possible values ‘X’ and ‘O’. The values ‘X’ always appear in form of rectangular islands and these islands are always row-wise and column-wise separated by at least one line of ‘O’s. Note that islands can only be diagonally adjacent. Count the number of islands in the given matrix.

Examples:

```mat[M][N] =  {{'O', 'O', 'O'},
{'X', 'X', 'O'},
{'X', 'X', 'O'},
{'O', 'O', 'X'},
{'O', 'O', 'X'},
{'X', 'X', 'O'}
};
Output: Number of islands is 3

mat[M][N] =  {{'X', 'O', 'O', 'O', 'O', 'O'},
{'X', 'O', 'X', 'X', 'X', 'X'},
{'O', 'O', 'O', 'O', 'O', 'O'},
{'X', 'X', 'X', 'O', 'X', 'X'},
{'X', 'X', 'X', 'O', 'X', 'X'},
{'O', 'O', 'O', 'O', 'X', 'X'},
};
Output: Number of islands is 4
```

We strongly recommend to minimize your browser and try this yourself first.

The idea is to count all top-leftmost corners of given matrix. We can check if a ‘X’ is top left or not by checking following conditions.
1) A ‘X’ is top of rectangle if the cell just above it is a ‘O’
2) A ‘X’ is leftmost of rectangle if the cell just left of it is a ‘O’

Note that we must check for both conditions as there may be more than one top cells and more than one leftmost cells in a rectangular island. Below is the implementation of above idea.

C/C++

```// A C++ program to count the number of rectangular
// islands where every island is separated by a line
#include<iostream>
using namespace std;

// Size of given matrix is M X N
#define M 6
#define N 3

// This function takes a matrix of 'X' and 'O'
// and returns the number of rectangular islands
// of 'X' where no two islands are row-wise or
// column-wise adjacent, the islands may be diagonaly
int countIslands(int mat[][N])
{
int count = 0; // Initialize result

// Traverse the input matrix
for (int i=0; i<M; i++)
{
for (int j=0; j<N; j++)
{
// If current cell is 'X', then check
// whether this is top-leftmost of a
// rectangle. If yes, then increment count
if (mat[i][j] == 'X')
{
if ((i == 0 || mat[i-1][j] == 'O') &&
(j == 0 || mat[i][j-1] == 'O'))
count++;
}
}
}

return count;
}

// Driver program to test above function
int main()
{
int mat[M][N] =  {{'O', 'O', 'O'},
{'X', 'X', 'O'},
{'X', 'X', 'O'},
{'O', 'O', 'X'},
{'O', 'O', 'X'},
{'X', 'X', 'O'}
};
cout << "Number of rectangular islands is "
<< countIslands(mat);
return 0;
}```

Java

```// A Java program to count the number of rectangular
// islands where every island is separated by a line
import java.io.*;

class islands
{
// This function takes a matrix of 'X' and 'O'
// and returns the number of rectangular islands
// of 'X' where no two islands are row-wise or
// column-wise adjacent, the islands may be diagonaly
static int countIslands(int mat[][], int m, int n)
{
// Initialize result
int count = 0;

// Traverse the input matrix
for (int i=0; i<m; i++)
{
for (int j=0; j<n; j++)
{
// If current cell is 'X', then check
// whether this is top-leftmost of a
// rectangle. If yes, then increment count
if (mat[i][j] == 'X')
{
if ((i == 0 || mat[i-1][j] == 'O') &&
(j == 0 || mat[i][j-1] == 'O'))
count++;
}
}
}

return count;
}

// Driver program
public static void main (String[] args)
{
// Size of given matrix is m X n
int m = 6;
int n = 3;
int mat[][] = {{'O', 'O', 'O'},
{'X', 'X', 'O'},
{'X', 'X', 'O'},
{'O', 'O', 'X'},
{'O', 'O', 'X'},
{'X', 'X', 'O'}
};
System.out.println("Number of rectangular islands is: "
+ countIslands(mat, m, n));
}
}

// Contributed by Pramod Kumar
```

Output:
`Number of rectangular islands is 3`

Time complexity of this solution is O(MN).

This article is contributed by Udit Gupta. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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