# Find perimeter of shapes formed with 1s in binary matrix

Given a matrix of N rows and M columns, consist of 0’s and 1’s. The task is to find the perimeter of subfigure consisting only 1’s in the matrix. Perimeter of single 1 is 4 as it can be covered from all 4 side. Perimeter of double 11 is 6.

```

|  1  |           |  1    1  |

```

Examples:

```Input : mat[][] =
{
1, 0,
1, 1,
}
Output : 6
Cell (1,0) and (1,1) making a L shape whose perimeter is 8.

Input :  mat[][] =
{
0, 1, 0, 0, 0,
1, 1, 1, 0, 0,
1, 0, 0, 0, 0
}
Output : 12

```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

The idea is to traverse the matrix, find all ones and find their contribution in perimeter. The maximum contribution of a 1 is four if it is surrounded by all 0s. The contribution reduces by one with 1 around it.

Algorithm for solving this problem:

1. Traverse the whole matrix and find the cell having value equal to 1.
2. Calculate the number of closed side for that cell and add, 4 – number of closed side to the total perimeter.

Below is C++ representation of this approach:

```// C++ program to find perimeter of area coverede by
// 1 in 2D matrix consisits of 0's and  1's.
#include<bits/stdc++.h>
using namespace std;
#define R 3
#define C 5

// Find the number of covered side for mat[i][j].
int numofneighbour(int mat[][C], int i, int j)
{
int count = 0;

// UP
if (i > 0 && mat[i - 1][j])
count++;

// LEFT
if (j > 0 && mat[i][j - 1])
count++;

// DOWN
if (i < R-1 && mat[i + 1][j])
count++;

// RIGHT
if (j < C-1 && mat[i][j + 1])
count++;

return count;
}

// Returns sum of perimeter of shapes formed with 1s
int findperimeter(int mat[R][C])
{
int perimeter = 0;

// Traversing the matrix and finding ones to
// calculate their contribution.
for (int i = 0; i < R; i++)
for (int j = 0; j < C; j++)
if (mat[i][j])
perimeter += (4 - numofneighbour(mat, i ,j));

return perimeter;
}

// Driven Program
int main()
{
int mat[R][C] =
{
0, 1, 0, 0, 0,
1, 1, 1, 0, 0,
1, 0, 0, 0, 0,
};

cout << findperimeter(mat) << endl;

return 0;
}
```

Output:

```12
```

Time Complexity : O(RC).

This article is contributed by Anuj Chauhan(anuj0503). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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