Find sum of all Boundary and Diagonal element of a Matrix

Given a 2D array arr[][] of order NxN, the task is to find the sum of all the elements present in both the diagonals and boundary elements of the given arr[][]. 
Examples: 
 

Input: arr[][] = { {1, 2, 3, 4}, {1, 2, 3, 4}, {1, 2, 3, 4}, {1, 2, 3, 4} } 
Output: 40 
Explanation: 
The Sum of elements on the boundary is 1 + 2 + 3 + 4 + 4 + 4 + 4 + 3 + 2 + 1 + 1 + 1 = 30. 
The Sum of elements on the diagonals which do not intersect with the boundary elements is 2 + 3 + 2 + 3 = 10. 
Therefore the required sum is 30 + 10 = 40.
Input: arr[][] = { {1, 2, 3}, {1, 2, 3}, {1, 2, 3}} 
Output: 18 
Explanation: 
The Sum of elements on the boundary is 1 + 2 + 3 + 3 + 3 + 2 + 1 + 1 = 16. 
The Sum of elements on the diagonals which do not intersect with the boundary elements is 2. 
Therefore the required sum is 16 + 2 = 18. 
 

Approach: 
 

1. Traverse the given 2D array with two loops, one for rows(say i) and another for columns(say j).
2. If i equals to j or (i + j) equals to (size of column - 1) then that element contributes to diagonals of the given 2D array.
3. If (i or j equals to 0) or (i or j equals to size of column - 1) then that element contributes to boundary elements of the given 2D array.
4. The sum of all the element satisfying above two conditions gives the required sum.

Below is the implementation of the above approach: 
 

// C++ implementation of the above approach

#include "bits/stdc++.h"
using namespace std;

const int N = 4;

// Function to find the sum of all diagonal
// and Boundary elements
void diagonalBoundarySum(int arr[N][N])
{

    int requiredSum = 0;

    // Traverse arr[][]
    // Loop from i to N-1 for rows
    for (int i = 0; i < N; i++) {

        // Loop from j = N-1 for columns
        for (int j = 0; j < N; j++) {

            // Condition for diagonal
            // elements
            if (i == j || (i + j) == N - 1) {
                requiredSum += arr[i][j];
            }

            // Condition for Boundary
            // elements
            else if (i == 0 || j == 0
                     || i == N - 1
                     || j == N - 1) {
                requiredSum += arr[i][j];
            }
        }
    }

    // Print the final Sum
    cout << requiredSum << endl;
}

// Driver Code
int main()
{
    int arr[][4] = { { 1, 2, 3, 4 },
                     { 1, 2, 3, 4 },
                     { 1, 2, 3, 4 },
                     { 1, 2, 3, 4 } };

    diagonalBoundarySum(arr);
    return 0;
}

// Java implementation of the above approach
import java.util.*;

class GFG{
    public static int N = 4;
    
    // Function to find the sum of all diagonal
    // and Boundary elements
    static void diagonalBoundarySum(int arr[][]){
        int requiredSum = 0;
        
        // Traverse arr[][]
        // Loop from i to N-1 for rows
        for (int i = 0; i < N; i++) {
    
            // Loop from j = N-1 for columns
            for (int j = 0; j < N; j++) {
    
                // Condition for diagonal
                // elements
                if (i == j || (i + j) == N - 1) {
                    requiredSum += arr[i][j];
                }
    
                // Condition for Boundary
                // elements
                else if (i == 0 || j == 0 || i == N - 1|| j == N - 1) {
                    requiredSum += arr[i][j];
                }
            }
        }
    
        // Print the final Sum
        System.out.println(requiredSum);
    }
    
    // Driver Code
    public static void main(String args[])
    {
        int arr[][] = { { 1, 2, 3, 4 },{ 1, 2, 3, 4 },
                        { 1, 2, 3, 4 },{ 1, 2, 3, 4 } };
    
        diagonalBoundarySum(arr);
        
    }
}

// This code is contributed by AbhiThakur

# Python implementation of the above approach

N = 4;

# Function to find the sum of all diagonal
# and Boundary elements
def diagonalBoundarySum(arr):
    requiredSum = 0;

    # Traverse arr
    # Loop from i to N-1 for rows
    for i in range(N):

        # Loop from j = N-1 for columns
        for j in range(N):

            # Condition for diagonal
            # elements
            if (i == j or (i + j) == N - 1):
                requiredSum += arr[i][j];
            
            # Condition for Boundary
            # elements
            elif(i == 0 or j == 0 or i == N - 1 or j == N - 1):
                requiredSum += arr[i][j];

    # Print the final Sum
    print(requiredSum);


# Driver Code
if __name__ == '__main__':
    arr = [[ 1, 2, 3, 4 ],
    [ 1, 2, 3, 4 ],
    [ 1, 2, 3, 4 ],
    [ 1, 2, 3, 4 ]];

    diagonalBoundarySum(arr);

# This code is contributed by 29AjayKumar

// C# implementation of the above approach
using System;

class GFG
{
    public static int N = 4;
    
    // Function to find the sum of all diagonal
    // and Boundary elements
    static void diagonalBoundarySum(int[, ] arr){
        int requiredSum = 0;
        
        // Traverse arr[][]
        // Loop from i to N-1 for rows
        for (int i = 0; i < N; i++) {
    
            // Loop from j = N-1 for columns
            for (int j = 0; j < N; j++) {
    
                // Condition for diagonal
                // elements
                if (i == j || (i + j) == N - 1) {
                    requiredSum += arr[i,j];
                }
    
                // Condition for Boundary
                // elements
                else if (i == 0 || j == 0 || i == N - 1|| j == N - 1) {
                    requiredSum += arr[i,j];
                }
            }
        }
    
        // Print the final Sum
        Console.WriteLine(requiredSum);
    }
    
    // Driver Code
    public static void Main()
    {
        int[, ] arr = { { 1, 2, 3, 4 },{ 1, 2, 3, 4 },{ 1, 2, 3, 4 },{ 1, 2, 3, 4 } };
    
        diagonalBoundarySum(arr);
        
    }
}

// This code is contributed by abhaysingh290895

<script>
// Java script  implementation of the above approach
let N = 4;
    
    // Function to find the sum of all diagonal
    // and Boundary elements
    function diagonalBoundarySum(arr){
        let requiredSum = 0;
        
        // Traverse arr[][]
        // Loop from i to N-1 for rows
        for (let i = 0; i < N; i++) {
    
            // Loop from j = N-1 for columns
            for (let j = 0; j < N; j++) {
    
                // Condition for diagonal
                // elements
                if (i == j || (i + j) == N - 1) {
                    requiredSum += arr[i][j];
                }
    
                // Condition for Boundary
                // elements
                else if (i == 0 || j == 0 || i == N - 1|| j == N - 1) {
                    requiredSum += arr[i][j];
                }
            }
        }
    
        // Print the final Sum
        document.write(requiredSum);
    }
    
    // Driver Code
    
        let arr = [[ 1, 2, 3, 4 ],[ 1, 2, 3, 4 ],
                        [1, 2, 3, 4 ],[ 1, 2, 3, 4 ]];
    
        diagonalBoundarySum(arr);
    
// contributed by sravan kumar
</script>

40

Time complexity: O(N*N) for given array of N*N size

Auxiliary space: O(1)

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Article Tags :

• Matrix
• C++ Programs
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• DSA
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