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

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:

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

Below is the implementation of this approach:

## C++

// 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; }

## Java

// Java program to find perimeter of area // coverede by 1 in 2D matrix consisits // of 0's and 1's class GFG { static final int R = 3; static final int C = 5; // Find the number of covered side // for mat[i][j]. static int numofneighbour(int mat[][], int i, int j) { int count = 0; // UP if (i > 0 && mat[i - 1][j] == 1) count++; // LEFT if (j > 0 && mat[i][j - 1] == 1) count++; // DOWN if (i < R - 1 && mat[i + 1][j] == 1) count++; // RIGHT if (j < C - 1 && mat[i][j + 1] == 1) count++; return count; } // Returns sum of perimeter of shapes // formed with 1s static int findperimeter(int mat[][]) { 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] == 1) perimeter += (4 - numofneighbour(mat, i, j)); return perimeter; } // Driver code public static void main(String[] args) { int mat[][] = {{0, 1, 0, 0, 0}, {1, 1, 1, 0, 0}, {1, 0, 0, 0, 0}}; System.out.println(findperimeter(mat)); } } // This code is contributed by Anant Agarwal.

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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