Efficiently compute sums of diagonals of a matrix

Given a 2D square matrix, find the sum of elements in Principal and Secondary diagonals. For example, consider the following 4 X 4 input matrix.

A00 A01 A02 A03
A10 A11 A12 A13
A20 A21 A22 A23
A30 A31 A32 A33

The primary diagonal is formed by the elements A00, A11, A22, A33. 

1. Condition for Principal Diagonal: The row-column condition is row = column. 
   The secondary diagonal is formed by the elements A03, A12, A21, A30.
2. Condition for Secondary Diagonal: The row-column condition is row = numberOfRows – column -1.

Examples :

Input : 
4
1 2 3 4
4 3 2 1
7 8 9 6
6 5 4 3
Output :
Principal Diagonal: 16
Secondary Diagonal: 20

Input :
3
1 1 1
1 1 1
1 1 1
Output :
Principal Diagonal: 3
Secondary Diagonal: 3

Method 1: In this method, we use two loops i.e. a loop for columns and a loop for rows and in the inner loop we check for the condition stated above:

Implementation:

C

// A simple C program to
// find sum of diagonals
#include <stdio.h>

const int M = 4;
const int N = 4;

void printDiagonalSums(int mat[M][N])
{
    int principal = 0, secondary = 0;
    for (int i = 0; i < M; i++)
    {
        for (int j = 0; j < N; j++)
        {
            // Condition for principal diagonal
            if (i == j)
                principal += mat[i][j];

            // Condition for secondary diagonal
            if ((i + j) == (N - 1))
                secondary += mat[i][j];
        }
    }

    printf("%s", "Principal Diagonal:");
    printf("%d\n", principal);
    printf("%s", "Secondary Diagonal:");
    printf("%d\n", secondary);
}

// Driver code
int main()
{
    int a[][4] = {{1, 2, 3, 4},
                  {5, 6, 7, 8},
                  {1, 2, 3, 4},
                  {5, 6, 7, 8}};
    printDiagonalSums(a);
    return 0;
}

C++

// A simple C++ program to find sum of diagonals
#include <bits/stdc++.h>
using namespace std;

const int MAX = 100;

void printDiagonalSums(int mat[][MAX], int n)
{
    int principal = 0, secondary = 0;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {

            // Condition for principal diagonal
            if (i == j)
                principal += mat[i][j];

            // Condition for secondary diagonal
            if ((i + j) == (n - 1))
                secondary += mat[i][j];
        }
    }

    cout << "Principal Diagonal:" << principal << endl;
    cout << "Secondary Diagonal:" << secondary << endl;
}

// Driver code
int main()
{
    int a[][MAX] = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 },
                    { 1, 2, 3, 4 }, { 5, 6, 7, 8 } };
    printDiagonalSums(a, 4);
    return 0;
}

Java

// A simple java program to find
// sum of diagonals
import java.io.*;

public class GFG {

    static void printDiagonalSums(int [][]mat,
                                         int n)
    {
        int principal = 0, secondary = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {

                // Condition for principal
                // diagonal
                if (i == j)
                    principal += mat[i][j];

                // Condition for secondary
                // diagonal
                if ((i + j) == (n - 1))
                    secondary += mat[i][j];
            }
        }

        System.out.println("Principal Diagonal:"
                                    + principal);

        System.out.println("Secondary Diagonal:"
                                    + secondary);
    }

    // Driver code
    static public void main (String[] args)
    {

        int [][]a = { { 1, 2, 3, 4 },
                      { 5, 6, 7, 8 },
                      { 1, 2, 3, 4 },
                      { 5, 6, 7, 8 } };

        printDiagonalSums(a, 4);
    }
}

// This code is contributed by vt_m.

Python3

# A simple Python program to
# find sum of diagonals
MAX = 100

def printDiagonalSums(mat, n):

    principal = 0
    secondary = 0;
    for i in range(0, n):
        for j in range(0, n):

            # Condition for principal diagonal
            if (i == j):
                principal += mat[i][j]

            # Condition for secondary diagonal
            if ((i + j) == (n - 1)):
                secondary += mat[i][j]

    print("Principal Diagonal:", principal)
    print("Secondary Diagonal:", secondary)

# Driver code
a = [[ 1, 2, 3, 4 ],
     [ 5, 6, 7, 8 ],
     [ 1, 2, 3, 4 ],
      [ 5, 6, 7, 8 ]]
printDiagonalSums(a, 4)

# This code is contributed
# by ihritik

C#

// A simple C# program to find sum
// of diagonals
using System;

public class GFG {

    static void printDiagonalSums(int [,]mat,
                                        int n)
    {
        int principal = 0, secondary = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {

                // Condition for principal
                // diagonal
                if (i == j)
                    principal += mat[i,j];

                // Condition for secondary
                // diagonal
                if ((i + j) == (n - 1))
                    secondary += mat[i,j];
            }
        }

        Console.WriteLine("Principal Diagonal:"
                                  + principal);

        Console.WriteLine("Secondary Diagonal:"
                                  + secondary);
    }

    // Driver code
    static public void Main ()
    {
        int [,]a = { { 1, 2, 3, 4 },
                     { 5, 6, 7, 8 },
                     { 1, 2, 3, 4 },
                     { 5, 6, 7, 8 } };

        printDiagonalSums(a, 4);
    }
}

// This code is contributed by vt_m.

PHP

<?php
// A simple PHP program to
// find sum of diagonals
$MAX = 100;

function printDiagonalSums($mat, $n)
{
    global $MAX;
    $principal = 0;
    $secondary = 0;
    for ($i = 0; $i < $n; $i++)
    {
        for ($j = 0; $j < $n; $j++)
        {

            // Condition for
            // principal diagonal
            if ($i == $j)
                $principal += $mat[$i][$j];

            // Condition for
            // secondary diagonal
            if (($i + $j) == ($n - 1))
                $secondary += $mat[$i][$j];
        }
    }

    echo "Principal Diagonal:" ,
               $principal ,"\n";
    echo "Secondary Diagonal:",
              $secondary ,"\n";
}

// Driver code
$a = array (array ( 1, 2, 3, 4 ),
            array ( 5, 6, 7, 8 ),
            array ( 1, 2, 3, 4 ),
            array ( 5, 6, 7, 8 ));
printDiagonalSums($a, 4);

// This code is contributed by ajit
?>

Javascript

<script>
// A simple Javascript program to find sum of diagonals

const MAX = 100;

void printDiagonalSums(mat, n)
{
    let principal = 0, secondary = 0;
    for (let i = 0; i < n; i++) {
        for (let j = 0; j < n; j++) {

            // Condition for principal diagonal
            if (i == j)
                principal += mat[i][j];

            // Condition for secondary diagonal
            if ((i + j) == (n - 1))
                secondary += mat[i][j];
        }
    }

    document.write("Principal Diagonal:" + principal + "<br>");
    document.write("Secondary Diagonal:" + secondary + "<br>");
}

// Driver code
    let a = [ [ 1, 2, 3, 4 ], [ 5, 6, 7, 8 ],
                    [ 1, 2, 3, 4 ], [ 5, 6, 7, 8 ] ];
    printDiagonalSums(a, 4);

// This code is contributed by subhammahato348.
</script>

Output

Principal Diagonal:18
Secondary Diagonal:18

Time Complexity: O(N*N), as we are using nested loops to traverse N*N times.
Auxiliary Space: O(1), as we are not using any extra space.

Method 2( Efficient Approach): In this method, we use one loop i.e. a loop for calculating the sum of both the principal and secondary diagonals: 

Implementation:

C

// An efficient C program to find
// sum of diagonals
#include <stdio.h>
#include <stdlib.h>

const int M = 4;
const int N = 4;

void printDiagonalSums(int mat[M][N])
{
    int principal = 0, secondary = 0;
    for (int i = 0; i < N; i++)
    {
        principal += mat[i][i];
        secondary += mat[i][N - i - 1];
    }

    printf("%s","Principal Diagonal:");
    printf("%d\n", principal);
    printf("%s", "Secondary Diagonal:");
    printf("%d\n", secondary);
}

// Driver code
int main()
{
    int a[4][4] = {{1, 2, 3, 4},
                   {5, 6, 7, 8},
                   {1, 2, 3, 4},
                   {5, 6, 7, 8}};
    printDiagonalSums(a);
    return 0;
}

C++

// An efficient C++ program to find sum of diagonals
#include <bits/stdc++.h>
using namespace std;

const int MAX = 100;

void printDiagonalSums(int mat[][MAX], int n)
{
    int principal = 0, secondary = 0;
    for (int i = 0; i < n; i++) {
        principal += mat[i][i];
        secondary += mat[i][n - i - 1];
    }

    cout << "Principal Diagonal:" << principal << endl;
    cout << "Secondary Diagonal:" << secondary << endl;
}

// Driver code
int main()
{
    int a[][MAX] = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 },
                     { 1, 2, 3, 4 }, { 5, 6, 7, 8 } };
    printDiagonalSums(a, 4);
    return 0;
}

Java

// An efficient java program to find
// sum of diagonals
import java.io.*;

public class GFG {

    static void printDiagonalSums(int [][]mat,
                                        int n)
    {
        int principal = 0, secondary = 0;
        for (int i = 0; i < n; i++) {
            principal += mat[i][i];
            secondary += mat[i][n - i - 1];
        }

        System.out.println("Principal Diagonal:"
                                   + principal);

        System.out.println("Secondary Diagonal:"
                                   + secondary);
    }

    // Driver code
    static public void main (String[] args)
    {
        int [][]a = { { 1, 2, 3, 4 },
                      { 5, 6, 7, 8 },
                      { 1, 2, 3, 4 },
                      { 5, 6, 7, 8 } };

        printDiagonalSums(a, 4);
    }
}

// This code is contributed by vt_m.

Python3

# A simple Python3 program to find
# sum of diagonals
MAX = 100

def printDiagonalSums(mat, n):

    principal = 0
    secondary = 0
    for i in range(0, n):
        principal += mat[i][i]
        secondary += mat[i][n - i - 1]

    print("Principal Diagonal:", principal)
    print("Secondary Diagonal:", secondary)

# Driver code
a = [[ 1, 2, 3, 4 ],
     [ 5, 6, 7, 8 ],
     [ 1, 2, 3, 4 ],
     [ 5, 6, 7, 8 ]]
printDiagonalSums(a, 4)

# This code is contributed
# by ihritik

C#

// An efficient C#program to find
// sum of diagonals
using System;

public class GFG {

    static void printDiagonalSums(int [,]mat,
                                       int n)
    {
        int principal = 0, secondary = 0;
        for (int i = 0; i < n; i++) {
            principal += mat[i,i];
            secondary += mat[i,n - i - 1];
        }

        Console.WriteLine("Principal Diagonal:"
                                  + principal);

        Console.WriteLine("Secondary Diagonal:"
                                  + secondary);
    }

    // Driver code
    static public void Main ()
    {
        int [,]a = { { 1, 2, 3, 4 },
                     { 5, 6, 7, 8 },
                     { 1, 2, 3, 4 },
                     { 5, 6, 7, 8 } };

        printDiagonalSums(a, 4);
    }
}

// This code is contributed by vt_m.

PHP

<?php
// An efficient PHP program
// to find sum of diagonals
$MAX = 100;

function printDiagonalSums($mat, $n)
{
    global $MAX;
    $principal = 0; $secondary = 0;
    for ($i = 0; $i < $n; $i++)
    {
        $principal += $mat[$i][$i];
        $secondary += $mat[$i][$n - $i - 1];
    }

    echo "Principal Diagonal:" ,
               $principal ,"\n";
    echo "Secondary Diagonal:" ,
               $secondary ,"\n";
}

// Driver Code
$a = array(array(1, 2, 3, 4),
           array(5, 6, 7, 8),
           array(1, 2, 3, 4),
           array(5, 6, 7, 8));
printDiagonalSums($a, 4);

// This code is contributed by aj_36
?>

Javascript

<script>

// An efficient Javascript  program to find
// sum of diagonals

function  printDiagonalSums(mat,n)
    {
        let principal = 0, secondary = 0;
        for (let i = 0; i < n; i++) {
            principal += mat[i][i];
            secondary += mat[i][n - i - 1];
        }

        document.write("Principal Diagonal:"
                                + principal+"<br>");

        document.write("Secondary Diagonal:"
                                + secondary);
    }

    // Driver code

        let a = [[ 1, 2, 3, 4 ],
                    [5, 6, 7, 8 ],
                    [ 1, 2, 3, 4 ],
                    [ 5, 6, 7, 8 ]];

        printDiagonalSums(a, 4);

// This code is contributed Bobby

</script>

Output

Principal Diagonal:18
Secondary Diagonal:18

Time Complexity: O(N), as we are using a loop to traverse N times.
Auxiliary Space: O(1), as we are not using any extra space.

This article is contributed by Mohak Agrawal. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.