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 diagonalsclass 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 diagonalsclass 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!
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