# C++ 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:

## C++

 `// C++ program to find the difference` `// between the sum of diagonal.` `#include ` `#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;` `}`

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.

## C++

 `// C++ program to find the difference` `// between the sum of diagonal.` `#include ` `#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 += arr[i][i];` `        ``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;` `}`

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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