# 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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 and (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];        # Prthe 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

Output:
40

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