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
perimeter

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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