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

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  • Last Updated : 27 Jan, 2022
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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 <bits/stdc++.h>
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)

Please refer complete article on Maximize sum of diagonal of a matrix by rotating all rows or all columns for more details!


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