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Maximum sum of elements in a diagonal parallel to the main diagonal of a given Matrix
  • Difficulty Level : Medium
  • Last Updated : 17 Mar, 2021

Give a square matrix mat[][] of dimensions N * N, the task is to find the maximum sum of elements present in the given matrix along the diagonals which are parallel to the main diagonal. Below is the image of the same.

Examples: 

Input: mat[][] = {{1, 2, 5, 7}, {2, 6, 7, 3}, {12, 3, 2, 4}, {3, 6, 9, 4}}
Output: 18
Explanation:
Sum of elements present in the diagonal having cells (2, 0) and (3, 1) is 12 + 6 = 18 which is maximum among all diagonals.

Input: mat[][] = {{5, 2, 5, 7}, {2, 5, 7, 3}, {12, 3, 5, 4}, {3, 6, 9, 5}}
Output: 18
Explanation:
Sum of elements present in the main diagonal having cells (0, 0), (1, 1), (2, 2) and (3, 3) is 5 + 5 + 5 + 5 = 20 which is maximum among all diagonals.

Approach: The idea is to traverse cells of each diagonal that is parallel to the main diagonal and observe that for any diagonal above the main diagonal starting at cell (x, y), it’s corresponding diagonal that is below the main diagonal will start at cell (y, x). For each diagonal, starting at cell (x, y) all its elements will be on cells (x + k, y + k) where 0 <= x + k, y + k < N. Follow the below steps to solve the problem:

  • Initialize a variable maxSum with 0 which will store the maximum diagonal sum.
  • Traverse the columns of 0th row from i over the range [0, N – 1].
  • Initialize variables sum1 and sum2 which will store the diagonal sums starting from the cell (row, col) and from the cell (col, row) respectively where r is 0 and c is col.
  • Increment both row and c by 1. Add mat[row][col] to sum1 and mat[col][row] to sum2 while row and col are smaller than N. Finally, update maxSum to store the maximum of maxSum, sum1, and sum2.
  • After traversing the given matrix, print the value maxSum as the maximum sum.

Below is the implementation of the above approach:



C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to return maximum diagonal
// sum that are parallel to main diagonal
int maxDiagonalSum(vector<vector<int> > arr, int N)
{
    // Initialize maxSum
    int maxSum = 0;
 
    // Traverse through the columns
    for (int i = 0; i < N; i++) {
 
        // Initialize r and c
        int row = 0, col = i;
 
        // Diagonal sums
        int sum1 = 0, sum2 = 0;
        while (col < N && row < N) {
            sum1 += arr[row][col];
            sum2 += arr[col][row];
            row++;
            col++;
        }
 
        // Update maxSum with
        // the maximum sum
        maxSum = max({ sum1, maxSum, sum2 });
    }
 
    // Return the maxSum
    return maxSum;
}
 
// Driver Code
int main()
{
    // Given matrix mat[][]
    vector<vector<int> > mat
        = { { 1, 2, 5, 7 },
            { 2, 6, 7, 3 },
            { 12, 3, 2, 4 },
            { 3, 6, 9, 4 } };
    int N = mat.size();
 
    // Function Call
    cout << maxDiagonalSum(mat, N);
 
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
 
class GFG{
   
// Function to return maximum diagonal
// sum that are parallel to main diagonal
static int maxDiagonalSum(int arr[][], int N)
{
     
    // Initialize maxSum
    int maxSum = 0;
 
    // Traverse through the columns
    for(int i = 0; i < N; i++)
    {
         
        // Initialize r and c
        int row = 0, col = i;
 
        // Diagonal sums
        int sum1 = 0, sum2 = 0;
        while (col < N && row < N)
        {
            sum1 += arr[row][col];
            sum2 += arr[col][row];
            row++;
            col++;
        }
 
        // Update maxSum with
        // the maximum sum
        maxSum = Math.max(maxSum,
                          Math.max(sum1,
                                   sum2));
    }
 
    // Return the maxSum
    return maxSum;
}
 
// Driver code
public static void main (String[] args)
{
     
    // Given matrix mat[][]
    int mat[][] = { { 1, 2, 5, 7 },
                    { 2, 6, 7, 3 },
                    { 12, 3, 2, 4 },
                    { 3, 6, 9, 4 } };
    int N = mat.length;
 
    // Function Call
    System.out.println(maxDiagonalSum(mat, N));
}
}
 
// This code is contributed by math_lover

Python3




# Python3 program for the above approach
 
# Function to return maximum diagonal
# sum that are parallel to main diagonal
def maxDiagonalSum(arr, N):
     
    # Initialize maxSum
    maxSum = 0
 
    # Traverse through the columns
    for i in range(N):
         
        # Initialize r and c
        row = 0
        col = i
 
        # Diagonal sums
        sum1 = 0
        sum2 = 0
         
        while col < N and row < N:
            sum1 += arr[row][col]
            sum2 += arr[col][row]
            row += 1
            col += 1
 
        # Update maxSum with
        # the maximum sum
        maxSum = max([ sum1, maxSum, sum2])
 
    # Return the maxSum
    return maxSum
 
# Driver Code
if __name__ == '__main__':
     
    # Given matrix mat[][]
    mat = [ [ 1, 2, 5, 7 ],
            [ 2, 6, 7, 3 ],
            [ 12, 3, 2, 4 ],
            [ 3, 6, 9, 4 ] ]
 
    N = len(mat)
 
    # Function Call
    print(maxDiagonalSum(mat, N))
 
# This code is contributed by mohit kumar 29

C#




// C# program for the
// above approach
using System;
class GFG{
   
// Function to return maximum
// diagonal sum that are parallel
// to main diagonal
static int maxDiagonalSum(int [,]arr,
                          int N)
{   
  // Initialize maxSum
  int maxSum = 0;
 
  // Traverse through the
  // columns
  for(int i = 0; i < N; i++)
  {
    // Initialize r and c
    int row = 0, col = i;
 
    // Diagonal sums
    int sum1 = 0, sum2 = 0;
    while (col < N && row < N)
    {
      sum1 += arr[row,col];
      sum2 += arr[col,row];
      row++;
      col++;
    }
 
    // Update maxSum with
    // the maximum sum
    maxSum = Math.Max(maxSum,
             Math.Max(sum1,
                      sum2));
  }
 
  // Return the maxSum
  return maxSum;
}
 
// Driver code
public static void Main(String[] args)
{   
  // Given matrix [,]mat
  int [,]mat = {{1, 2, 5, 7},
                {2, 6, 7, 3},
                {12, 3, 2, 4},
                {3, 6, 9, 4}};
  int N = mat.GetLength(0);
 
  // Function Call
  Console.WriteLine(maxDiagonalSum(mat, N));
}
}
 
// This code is contributed by gauravrajput1

Javascript




<script>
// javascript program for the above approach
 
// Function to return maximum diagonal
// sum that are parallel to main diagonal
function maxDiagonalSum( arr,  N)
{
    // Initialize maxSum
    let maxSum = 0;
 
    // Traverse through the columns
    for (let i = 0; i < N; i++) {
 
        // Initialize r and c
        let row = 0, col = i;
 
        // Diagonal sums
        let sum1 = 0, sum2 = 0;
        while (col < N && row < N) {
            sum1 += arr[row][col];
            sum2 += arr[col][row];
            row++;
            col++;
        }
 
        // Update maxSum with
        // the maximum sum
        maxSum = Math.max(Math.max(sum1, maxSum), sum2 );
    }
 
    // Return the maxSum
    return maxSum;
}
 
// Driver Code
 
    // Given matrix mat[][]
    let mat
        = [[ 1, 2, 5, 7 ],
            [ 2, 6, 7, 3 ],
            [ 12, 3, 2, 4 ],
            [ 3, 6, 9, 4 ]];
    let N = mat[0].length;
 
    // Function Call
     document.write(maxDiagonalSum(mat, N));
 
// This code is contributed by todaysgaurav
</script>
Output
18

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

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