Find difference between sums of two diagonals

Given a matrix of n X n. The task is to calculate the absolute difference between the sums of its diagonal.

Examples:

Input : mat[][] = 11 2 4
                   4 5 6
                  10 8 -12 
Output : 15
Sum of primary diagonal = 11 + 5 + (-12) = 4.
Sum of secondary diagonal = 4 + 5 + 10 = 19.
Difference = |19 - 4| = 15.


Input : mat[][] = 10 2
                   4 5
Output : 9

Calculate the sums across the two diagonals of a square matrix. Along the first diagonal of the matrix, row index = column index i.e mat[i][j] lies on the first diagonal if i = j. Along the other diagonal, row index = n – 1 – column index i.e mat[i][j] lies on the second diagonal if i = n-1-j. By using two loops we traverse the entire matrix and calculate the sum across the diagonals of the matrix.

Below is the implementation of this approach: 

C++

// C++ program to find the difference
// between the sum of diagonal.
#include <bits/stdc++.h>
#define MAX 100
using namespace std;

int difference(int arr[][MAX], int n)
{
    // Initialize sums of diagonals
    int d1 = 0, d2 = 0;

    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < n; j++)
        {
            // finding sum of primary diagonal
            if (i == j)
                d1 += arr[i][j];

            // finding sum of secondary diagonal
            if (i == n - j - 1)
                d2 += arr[i][j];
        }
    }

    // Absolute difference of the sums
    // across the diagonals
    return abs(d1 - d2);
}

// Driven Program
int main()
{
    int n = 3;

    int arr[][MAX] =
    {
        {11, 2, 4},
        {4 , 5, 6},
        {10, 8, -12}
    };

    cout << difference(arr, n);
    return 0;
}

Java

// JAVA Code for Find difference between sums
// of two diagonals
import java.io.*;
class GFG {

    public static int difference(int arr[][], int n)
    {
        // Initialize sums of diagonals
        int d1 = 0, d2 = 0;

        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < n; j++)
            {
                // finding sum of primary diagonal
                if (i == j)
                    d1 += arr[i][j];

                // finding sum of secondary diagonal
                if (i == n - j - 1)
                    d2 += arr[i][j];
            }
        }

        // Absolute difference of the sums
        // across the diagonals
        return Math.abs(d1 - d2);
    }

    /* Driver program to test above function */
    public static void main(String[] args)
    {
        int n = 3;

        int arr[][] =
        {
            {11, 2, 4},
            {4 , 5, 6},
            {10, 8, -12}
        };

        System.out.print(difference(arr, n));

    }
  }
// This code is contributed by Arnav Kr. Mandal.

Python3

# Python3 program to find the difference
# between the sum of diagonal.
def difference(arr, n):

    # Initialize sums of diagonals
    d1 = 0
    d2 = 0

    for i in range(0, n):

        for j in range(0, n):

            # finding sum of primary diagonal
            if (i == j):
                d1 += arr[i][j]

            # finding sum of secondary diagonal
            if (i == n - j - 1):
                d2 += arr[i][j]

    # Absolute difference of the sums
    # across the diagonals
    return abs(d1 - d2);

# Driver Code
n = 3

arr = [[11, 2, 4],
       [4 , 5, 6],
       [10, 8, -12]]

print(difference(arr, n))

# This code is contributed
# by ihritik

C#

// C# Code for find difference between
// sums of two diagonals
using System;

public class GFG
{

    // Function to calculate difference
    public static int difference(int[,] arr,
                                 int n)
    {

        // Initialize sums of diagonals
        int d1 = 0, d2 = 0;

        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < n; j++)
            {

                // finding sum of primary diagonal
                if (i == j)
                    d1 += arr[i, j];

                // finding sum of secondary diagonal
                if (i == n - j - 1)
                    d2 += arr[i, j];
            }
        }

        // Absolute difference of the
        // sums across the diagonals
        return Math.Abs(d1 - d2);
    }

    // Driver Code
    public static void Main()
    {
        int n = 3;

        int[,] arr ={{11, 2, 4},
                     {4 , 5, 6},
                     {10, 8, -12}};

        Console.Write(difference(arr, n));

    }
}

// This code is contributed by shiv_bhakt.

PHP

<?php
// PHP program to find the difference
// between the sum of diagonal.

function difference($arr, $n)
{

    // Initialize sums of diagonals
    $d1 = 0; $d2 = 0;

    for ($i = 0; $i < $n; $i++)
    {
        for ($j = 0; $j < $n; $j++)
        {

            // finding sum of
            // primary diagonal
            if ($i == $j)
                $d1 += $arr[$i][$j];

            // finding sum of
            // secondary diagonal
            if ($i == $n - $j - 1)
                $d2 += $arr[$i][$j];
        }
    }

    // Absolute difference of the sums
    // across the diagonals
    return abs($d1 - $d2);
}

// Driver Code
{
    $n = 3;

    $arr = array(array(11, 2, 4),
                 array(4 , 5, 6),
                 array(10, 8, -12));

    echo difference($arr, $n);
    return 0;
}

// This code is contributed by nitin mittal.
?>

Javascript

<script>

// Javascript Code for Find difference between sums
// of two diagonals

function difference(arr,n)
    {
        // Initialize sums of diagonals
        let d1 = 0, d2 = 0;

        for (let i = 0; i < n; i++)
        {
            for (let j = 0; j < n; j++)
            {
                // finding sum of primary diagonal
                if (i == j)
                    d1 += arr[i][j];

                // finding sum of secondary diagonal
                if (i == n - j - 1)
                    d2 += arr[i][j];
            }
        }

        // Absolute difference of the sums
        // across the diagonals
        return Math.abs(d1 - d2);
    }

    /* Driver program to test above function */

        let n = 3;

        let arr =
        [
            [11, 2, 4],
            [4 , 5, 6],
            [10, 8, -12]
        ];

        document.write(difference(arr, n));

// This code is contributed Bobby

</script>

Output

15

Time complexity: O(n*n)
Auxiliary Space: O(1) using constant space to initialize diagonal sums, since no extra space has been taken.

We can optimize the above solution to work in O(n) using the patterns present in indexes of cells. 

C++

// C++ program to find the difference
// between the sum of diagonal.
#include <bits/stdc++.h>
#define MAX 100
using namespace std;

int difference(int arr[][MAX], int n)
{
    // Initialize sums of diagonals
    int d1 = 0, d2 = 0;

    for (int i = 0; i < n; i++)
    {
          // d1 store the sum of diagonal from
          // top-left to bottom-right
        d1 += arr[i][i];

          // d2 store the sum of diagonal from
        // top-right to bottom-left
        d2 += arr[i][n-i-1];
    }

    // Absolute difference of the sums
    // across the diagonals
    return abs(d1 - d2);
}

// Driven Program
int main()
{
    int n = 3;

    int arr[][MAX] =
    {
        {11, 2, 4},
        {4 , 5, 6},
        {10, 8, -12}
    };

    cout << difference(arr, n);
    return 0;
}

Java

// JAVA Code for Find difference between sums
// of two diagonals
import java.io.*;
class GFG {

    public static int difference(int arr[][], int n)
    {
        // Initialize sums of diagonals
        int d1 = 0, d2 = 0;

        for (int i = 0; i < n; i++)
        {
            d1 += arr[i][i];
            d2 += arr[i][n-i-1];
        }

        // Absolute difference of the sums
        // across the diagonals
        return Math.abs(d1 - d2);
    }

    /* Driver program to test above function */
    public static void main(String[] args)
    {
        int n = 3;

        int arr[][] =
        {
            {11, 2, 4},
            {4 , 5, 6},
            {10, 8, -12}
        };

        System.out.print(difference(arr, n));

    }
  }
// This code is contributed by Arnav Kr. Mandal.

Python3

# Python3 program to find the difference
# between the sum of diagonal.
def difference(arr, n):

    # Initialize sums of diagonals
    d1 = 0
    d2 = 0

    for i in range(0, n):
        d1 = d1 + arr[i][i]
        d2 = d2 + arr[i][n - i - 1]

    # Absolute difference of the sums
    # across the diagonals
    return abs(d1 - d2)

# Driver Code
n = 3

arr = [[11, 2, 4],
       [4 , 5, 6],
       [10, 8, -12]]

print(difference(arr, n))

# This code is contributed
# by ihritik

C#

// C# Code for find difference between
// sums of two diagonals
using System;

public class GFG

{

    //Function to find difference
    public static int difference(int[,] arr,
                                 int n)
    {

        // Initialize sums of diagonals
        int d1 = 0, d2 = 0;

        for (int i = 0; i < n; i++)
        {
            d1 += arr[i, i];
            d2 += arr[i, n - i - 1];
        }

        // Absolute difference of the sums
        // across the diagonals
        return Math.Abs(d1 - d2);
    }

    // Driver Code
    public static void Main()
    {
        int n = 3;

        int[,] arr ={{11, 2, 4},
                     {4 , 5, 6},
                     {10, 8, -12}};

        Console.Write(difference(arr, n));

    }
}

// This code is contributed by shiv_bhakt.

PHP

<?php
// PHP program to find the difference
// between the sum of diagonal.

function difference($arr, $n)
{

    // Initialize sums of diagonals
    $d1 = 0; $d2 = 0;

    for ($i = 0; $i < $n; $i++)
    {
        $d1 += $arr[$i][$i];
        $d2 += $arr[$i][$n-$i-1];
    }

    // Absolute difference of the sums
    // across the diagonals
    return abs($d1 - $d2);
}

// Driver Code
{
    $n = 3;

    $arr =array(array(11, 2, 4),
                array(4, 5, 6),
                array(10, 8, -12));

    echo difference($arr, $n);
    return 0;
}

// This code is contributed by nitin mittal.
?>

Javascript

<script>
// JAVA SCRIPT  Code for Find difference between sums
// of two diagonals

function difference(arr,n)
    {

        // Initialize sums of diagonals
        let d1 = 0, d2 = 0;

        for (let i = 0; i < n; i++)
        {
            d1 += arr[i][i];
            d2 += arr[i][n-i-1];
        }

        // Absolute difference of the sums
        // across the diagonals
        return Math.abs(d1 - d2);
    }

    /* Driver program to test above function */

        let n = 3;
        let arr = [[11, 2, 4],
            [4 , 5, 6],
            [10, 8, -12]];

        document.write(difference(arr, n));


// This code is contributed by sravan kumar Gottumukkala
</script>

Output

15

Time complexity : O(n)
Auxiliary Space: O(1) using constant space to initialize diagonal sums

This article is contributed by Anuj Chauhan. 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.