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Find perimeter of shapes formed with 1s in binary matrix

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  • Difficulty Level : Easy
  • Last Updated : 06 Jul, 2022

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 : 8
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 of shapes formed with 1s in binary matrix

Recommended Practice

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 the implementation of this approach: 

C++




// C++ program to find perimeter of area covered by
// 1 in 2D matrix consists 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
// covered by 1 in 2D matrix consists
// 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.


Python3




# Python3 program to find perimeter of area
# covered by 1 in 2D matrix consists of 0's and 1's.
 
R = 3
C = 5
 
# Find the number of covered side for mat[i][j].
def numofneighbour(mat, i, j):
 
    count = 0;
 
    # UP
    if (i > 0 and mat[i - 1][j]):
        count+= 1;
 
    # LEFT
    if (j > 0 and mat[i][j - 1]):
        count+= 1;
 
    # DOWN
    if (i < R-1 and mat[i + 1][j]):
        count+= 1
 
    # RIGHT
    if (j < C-1 and mat[i][j + 1]):
        count+= 1;
 
    return count;
 
# Returns sum of perimeter of shapes formed with 1s
def findperimeter(mat):
 
    perimeter = 0;
 
    # Traversing the matrix and finding ones to
    # calculate their contribution.
    for i in range(0, R):
        for j in range(0, C):
            if (mat[i][j]):
                perimeter += (4 - numofneighbour(mat, i, j));
 
    return perimeter;
 
# Driver Code
mat = [ [0, 1, 0, 0, 0],
        [1, 1, 1, 0, 0],
        [1, 0, 0, 0, 0] ]
 
print(findperimeter(mat), end="\n");
 
# This code is contributed by Akanksha Rai


C#




using System;
 
// C# program to find perimeter of area
// covered by 1 in 2D matrix consists 
// of 0's and 1's
public class GFG
{
 
    public  const int R = 3;
    public const int C = 5;
 
    // Find the number of covered side 
    // for mat[i][j].
    public 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
    public 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 = new int[][]
        {
            new int[] {0, 1, 0, 0, 0},
            new int[] {1, 1, 1, 0, 0},
            new int[] {1, 0, 0, 0, 0}
        };
 
        Console.WriteLine(findperimeter(mat));
    }
}
 
// This code is contributed by Shrikant13


PHP




<?php
// PHP program to find perimeter of area
// covered by 1 in 2D matrix consists
// of 0's and 1's.
$R = 3;
$C = 5;
 
// Find the number of covered side
// for mat[i][j].
function numofneighbour($mat, $i, $j)
{
    global $R;
    global $C;
    $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
function findperimeter($mat)
{
    global $R;
    global $C;
    $perimeter = 0;
 
    // Traversing the matrix and finding ones
    // to calculate their contribution.
    for ($i = 0; $i < $R; $i++)
        for ( $j = 0; $j < $C; $j++)
            if ($mat[$i][$j])
                $perimeter += (4 -
                numofneighbour($mat, $i, $j));
 
    return $perimeter;
}
 
// Driver Code
$mat = array(array(0, 1, 0, 0, 0),
             array(1, 1, 1, 0, 0),
             array(1, 0, 0, 0, 0));
 
echo findperimeter($mat), "\n";
 
// This code is contributed by Sach_Code
?>


Javascript




<script>
 
// JavaScript program to find perimeter of area
// covered by 1 in 2D matrix consists
// of 0's and 1's
let R = 3;
let C = 5;
   
// Find the number of covered side
// for mat[i][j].
function numofneighbour(mat, i, j)
{
    let 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
function findperimeter(mat)
{
    let perimeter = 0;
   
    // Traversing the matrix and
    // finding ones to calculate
    // their contribution.
    for(let i = 0; i < R; i++)
        for(let j = 0; j < C; j++)
            if (mat[i][j] == 1)
                perimeter += (4 -
                numofneighbour(mat, i, j));
   
    return perimeter;
}
 
// Driver Code
let mat = [ [ 0, 1, 0, 0, 0 ],
            [ 1, 1, 1, 0, 0 ],
            [ 1, 0, 0, 0, 0 ] ];
              
document.write(findperimeter(mat));
 
// This code is contributed by souravghosh0416
 
</script>


Output

12

Time Complexity : O(R x C).
Auxiliary Space: O(1)

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


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