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Maximize sum of diagonal of a matrix by rotating all rows or all columns
  • Difficulty Level : Easy
  • Last Updated : 23 Apr, 2021

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;
}

Java




// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
 
static int N = 3;
  
// Function to find maximum sum of
// diagonal elements of matrix by
// rotating either rows or columns
static int findMaximumDiagonalSumOMatrixf(int A[][])
{
     
    // Stores maximum diagonal sum of elements
    // of matrix by rotating rows or columns
    int maxDiagonalSum = Integer.MIN_VALUE;
     
    // 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 = Math.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 = Math.max(maxDiagonalSum,
                                  curr);
    }
    return maxDiagonalSum;
}
  
// Driver Code
public static void main(String[] args)
{
    int[][] mat = { { 1, 1, 2 },
                    { 2, 1, 2 },
                    { 1, 2, 2 } };
      
    System.out.println(
        findMaximumDiagonalSumOMatrixf(mat));
}
}
 
// This code is contributed by susmitakundugoaldanga

Python3




# Python3 program to implement
# the above approach
import sys
 
N = 3
 
# Function to find maximum sum of diagonal
# elements of matrix by rotating either
# rows or columns
def findMaximumDiagonalSumOMatrixf(A):
     
    # Stores maximum diagonal sum of elements
    # of matrix by rotating rows or columns
    maxDiagonalSum = -sys.maxsize - 1
 
    # Rotate all the columns by an integer
    # in the range [0, N - 1]
    for i in range(N):     
 
        # Stores sum of diagonal elements
        # of the matrix
        curr = 0
         
        # Calculate sum of diagonal
        # elements of the matrix
        for j in range(N):
             
            # 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 i in range(N):
         
        # Stores sum of diagonal elements
        # of the matrix
        curr = 0
         
        # Calculate sum of diagonal
        # elements of the matrix
        for j in range(N):         
             
            # Update curr
            curr += A[(i + j) % N][j]
         
        # Update maxDiagonalSum
        maxDiagonalSum = max(maxDiagonalSum,
                             curr)
                              
    return maxDiagonalSum
 
# Driver code
if __name__ == "__main__":
     
    mat = [ [ 1, 1, 2 ],
            [ 2, 1, 2 ],
            [ 1, 2, 2 ] ]
     
    print(findMaximumDiagonalSumOMatrixf(mat))
     
# This code is contributed by chitranayal

C#




// C# program to implement
// the above approach 
using System;
   
class GFG{
   
static int N = 3;
   
// Function to find maximum sum of
// diagonal elements of matrix by
// rotating either rows or columns
static int findMaximumDiagonalSumOMatrixf(int[,] A)
{
     
    // Stores maximum diagonal sum of elements
    // of matrix by rotating rows or columns
    int maxDiagonalSum = Int32.MinValue;
     
    // 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 = Math.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 = Math.Max(maxDiagonalSum,
                                  curr);
    }
    return maxDiagonalSum;
}
   
// Driver Code
public static void Main()
{
    int[,] mat = { { 1, 1, 2 },
                   { 2, 1, 2 },
                   { 1, 2, 2 } };
       
    Console.Write(findMaximumDiagonalSumOMatrixf(mat));
}
}
 
// This code is contributed by code_hunt

Javascript




<script>
 
// Javascript program to implement
// the above approach
let N = 3;
   
// Function to find maximum sum of
// diagonal elements of matrix by
// rotating either rows or columns
function findMaximumDiagonalSumOMatrixf(A)
{
      
    // Stores maximum diagonal sum of elements
    // of matrix by rotating rows or columns
    let maxDiagonalSum = Number.MIN_VALUE;
      
    // Rotate all the columns by an leteger
    // in the range [0, N - 1]
    for(let i = 0; i < N; i++)
    {
          
        // Stores sum of diagonal elements
        // of the matrix
        let curr = 0;
          
        // Calculate sum of diagonal
        // elements of the matrix
        for(let j = 0; j < N; j++)
        {
              
            // Update curr
            curr += A[j][(i + j) % N];
        }
           
        // Update maxDiagonalSum
        maxDiagonalSum = Math.max(maxDiagonalSum,
                                  curr);
    }
      
    // Rotate all the rows by an leteger
    // in the range [0, N - 1]
    for(let i = 0; i < N; i++)
    {
          
        // Stores sum of diagonal elements
        // of the matrix
        let curr = 0;
          
        // Calculate sum of diagonal
        // elements of the matrix
        for(let j = 0; j < N; j++)
        {
              
            // Update curr
            curr += A[(i + j) % N][j];
        }
          
        // Update maxDiagonalSum
        maxDiagonalSum = Math.max(maxDiagonalSum,
                                  curr);
    }
    return maxDiagonalSum;
}
   
    // Driver Code
    let mat = [[ 1, 1, 2 ],
                    [ 2, 1, 2 ],
                    [ 1, 2, 2 ]];
       
    document.write(
        findMaximumDiagonalSumOMatrixf(mat));
 
// This code is contributed by souravghosh0416.
</script>
Output: 
6

 

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

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