Find perimeter of shapes formed with 1s in binary matrix Last Updated : 23 Jul, 2025 Comments Improve Suggest changes Like Article Like Report Try it on GfG Practice Given a matrix of N rows and M columns, consisting of 0's and 1's. The task is to find the perimeter of the subfigure consisting of only 1s in the matrix. The perimeter of a figure is given by the total number of horizontal and vertical linings formed by the boundary 1s of the subfigure. Each line is considered to have a length of 1 unit.Note: There exists only one subfigure consisting of 1s Examples: Input : mat[][] = {{1,0},{1,1}}Output : 8Explanation: Example 1Input : mat[][] = {{0, 1, 0, 0, 0},{1, 1, 1, 0, 0},{1, 0, 0, 0, 0}}Output : 12Explanation: Example 2Recommended PracticeFind perimeter of shapesTry It!Approach: The idea is to traverse the matrix, find all ones and find their contribution in perimeter as per the below cases: Case 1 (4 adjacent 1s) : contribution = 0Case 2 (3 adjacent 1s) : contribution = 1Case 3 (2 adjacent 1s) : contribution = 2Case 4 (1 adjacent 1s) : contribution = 3Case 5 (0 adjacent 1s) : contribution = 4 Finally return the sum of contribution of each 1 in the matrix as the answer. Traverse the whole matrix and find the cell having value equal to 1.Calculate the number of adjacent 1s for that cell and add, 4 - number of adjacent 1s 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 import java.io.*; 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 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> 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 ?> Output12 Time Complexity: O(R x C).Auxiliary Space: O(1), since no extra space has been taken. 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