# C++ Program to Maximize sum of diagonal of a matrix by rotating all rows or all columns

• Last Updated : 27 Jan, 2022

Given a square matrix, mat[][] of dimensions N * N, the task is find the maximum sum of diagonal elements possible from the given matrix by rotating either all the rows or all the columns of the matrix by a positive integer.

Examples:

Input: mat[][] = { { 1, 1, 2 }, { 2, 1, 2 }, { 1, 2, 2 } }
Output:
Explanation:
Rotating all the columns of matrix by 1 modifies mat[][] to { {2, 1, 2}, {1, 2, 2}, {1, 1, 2} }.
Therefore, the sum of diagonal elements of the matrix = 2 + 2 + 2 = 6 which is the maximum possible.

Input: A[][] = { { -1, 2 }, { -1, 3 } }
Output: 2

Approach: The idea is to rotate all the rows and columns of the matrix in all possible ways and calculate the maximum sum obtained. Follow the steps to solve the problem:

• Initialize a variable, say maxDiagonalSum to store the maximum possible sum of diagonal elements the matrix by rotating all the rows or columns of the matrix.
• Rotate all the rows of the matrix by a positive integer in the range [0, N – 1] and update the value of maxDiagonalSum.
• Rotate all the columns of the matrix by a positive integer in the range [0, N – 1] and update the value of maxDiagonalSum.
• Finally, print the value of maxDiagonalSum.

Below is the implementation of the above approach:

## C++

 `// C++ program to implement``// the above approach`` ` `#include ``using` `namespace` `std;``#define N 3`` ` ` ` `// Function to find maximum sum of diagonal elements ``// of matrix by rotating either rows or columns``int` `findMaximumDiagonalSumOMatrixf(``int` `A[][N])``{``     ` `     ` `    ``// Stores maximum diagonal sum of elements``    ``// of matrix by rotating rows or columns``    ``int` `maxDiagonalSum = INT_MIN;``     ` ` ` `    ``// Rotate all the columns by an integer``    ``// in the range [0, N - 1]``    ``for` `(``int` `i = 0; i < N; i++) {``         ` ` ` `        ``// Stores sum of diagonal elements``        ``// of the matrix``        ``int` `curr = 0;``         ` ` ` `        ``// Calculate sum of diagonal ``        ``// elements of the matrix``        ``for` `(``int` `j = 0; j < N; j++) {``             ` ` ` `            ``// Update curr``            ``curr += A[j][(i + j) % N];``        ``}``         ` `         ` `        ``// Update maxDiagonalSum``        ``maxDiagonalSum = max(maxDiagonalSum, ``                                      ``curr);``    ``}`` ` ` ` `    ``// Rotate all the rows by an integer``    ``// in the range [0, N - 1]``    ``for` `(``int` `i = 0; i < N; i++) {``         ` ` ` `        ``// Stores sum of diagonal elements``        ``// of the matrix``        ``int` `curr = 0;``         ` ` ` `        ``// Calculate sum of diagonal ``        ``// elements of the matrix``        ``for` `(``int` `j = 0; j < N; j++) {``             ` ` ` `            ``// Update curr``            ``curr += A[(i + j) % N][j];``        ``}``         ` `         ` `        ``// Update maxDiagonalSum``        ``maxDiagonalSum = max(maxDiagonalSum, ``                                      ``curr);``    ``}`` ` `       ` `    ``return` `maxDiagonalSum;``}`` ` ` ` `// Driver code``int` `main()``{``     ` `    ``int` `mat[N][N] = { { 1, 1, 2 }, ``                    ``{ 2, 1, 2 }, ``                    ``{ 1, 2, 2 } };``     ` `    ``cout<< findMaximumDiagonalSumOMatrixf(mat);``    ``return` `0;``}`

Output:

`6`

Time Complexity: O(N2)
Auxiliary Space: O(1)

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