Tutorialspoint.dev

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 : 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



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 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.

Python 3

# Python 3 program to find perimeter of area 
# covered by 1 in 2D matrix consisits 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=" ");
  
# This code is contributed by Akanksha Rai

C#

using System;
  
// C# program to find perimeter of area 
// coverede by 1 in 2D matrix consisits  
// 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 consisits
// 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), " "
  
// This code is contributed by Sach_Code
?>


Output:

12

Time Complexity : O(RC).

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



This article is attributed to GeeksforGeeks.org

leave a comment

code

0 Comments

load comments

Subscribe to Our Newsletter