Java Program to 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 primary diagonal = 4 + 5 + 10 = 19.
Difference = |19 - 4| = 15.


Input : mat[][] = 10 2
                   4 5
Output : 7

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: 

Java

// JAVA Code for Find difference between sums
// of two diagonals
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.

Output:  

15

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

Java

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

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.

Output:  

15

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.

Please refer complete article on Find difference between sums of two diagonals for more details!