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
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; } |
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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. |
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Python 3
# Python 3 program to find perimeter of area # covered by 1 in 2D matrix consisits of 0's and 1's. R = 3C = 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 |
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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 |
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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), "\n"; // This code is contributed by Sach_Code ?> |
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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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