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Construct a matrix with sum equal to the sum of diagonal elements
  • Last Updated : 04 Aug, 2020

Given an integer N, the task is to construct a matrix of size N2 using positive and negative integers and excluding 0, such that the sum of the matrix is equal to the sum of the diagonal of the matrix.
Examples: 
 

Input: N = 2 
Output: 
1 -2 
2 4 
Explanation: 
Diagonal sum = (1 + 4) = 5 
Matrix sum = (1 – 2 + 2 + 4) = 5 
Input: N = 5 
Output: 
1 2 3 5 10 
3 1 4 -9 1 
-19 6 1 5 -8 
4 -7 2 1 12 
-17 1 1 1 1 
Explanation: 
Diagonal sum = (1 + 1 + 1 + 1 + 1) = 5 
Matrix sum = 5 
 

 

Approach: 
The approach to solving the problem is to traverse all indices of the matrix and print a positive element(say y) at the N diagonal positions and equally distribute a single-valued positive and negative integer(say x and -x) in the remaining N2 – N positions.
Below is the implementation of the above approach:
 

C++




// C++ program to implement
// the above approach
#include <bits/stdc++.h> 
using namespace std;
  
// Function to construct matrix with
// diagonal sum equal to matrix sum
void constructmatrix(int N)
{
    bool check = true;
  
    for (int i = 0; i < N; i++) {
  
        for (int j = 0; j < N; j++) {
  
            // If diagonal position
            if (i == j) {
                cout << 1 << " ";
            }
  
            else if (check) {
  
                // Positve element
                cout << 2 << " ";
                check = false;
            }
            else {
  
                // Negative element
                cout << -2 << " ";
                check = true;
            }
        }
  
        cout << endl;
    }
}
  
// Driver Code
int main()
{
    int N = 5;
  
    constructmatrix(5);
  
    return 0;
}

Java




// Java program to implement
// the above approach
public class Main {
  
    // Function to construct matrix with
    // diagonal sum equal to matrix sum
    public static void constructmatrix(int N)
    {
        boolean check = true;
  
        for (int i = 0; i < N; i++) {
  
            for (int j = 0; j < N; j++) {
  
                // If diagonal position
                if (i == j) {
                    System.out.print("1 ");
                }
                else if (check) {
  
                    // Positve element
                    System.out.print("2 ");
                    check = false;
                }
                else {
                    // Negative element
                    System.out.print("-2 ");
                    check = true;
                }
            }
  
            System.out.println();
        }
    }
  
    // Driver Code
    public static void main(String[] args)
    {
        int N = 5;
  
        constructmatrix(5);
    }
}

Python3




# Python3 program to implement 
# the above approach 
  
# Function to construct matrix with 
# diagonal sum equal to matrix sum 
def constructmatrix(N): 
      
    check = bool(True
  
    for i in range(N):
        for j in range(N):
  
            # If diagonal position 
            if (i == j): 
                print(1, end = " ")
  
            elif (check): 
  
                # Positve element 
                print(2, end = " ")
                check = bool(False
                  
            else:
  
                # Negative element 
                print(-2, end = " ")
                check = bool(True
  
        print()
  
# Driver code
N = 5
constructmatrix(5)
  
# This code is contributed by divyeshrabadiya07

C#




// C# program to implement
// the above approach
using System;
  
class GFG{
      
// Function to construct matrix with
// diagonal sum equal to matrix sum
public static void constructmatrix(int N)
{
    bool check = true;
  
    for(int i = 0; i < N; i++)
    {
        for(int j = 0; j < N; j++)
        {
              
            // If diagonal position
            if (i == j)
            {
                Console.Write("1 ");
            }
            else if (check)
            {
                  
                // Positve element
                Console.Write("2 ");
                check = false;
            }
            else
            {
                  
                // Negative element
                Console.Write("-2 ");
                check = true;
            }
        }
        Console.WriteLine();
    }
}
  
// Driver Code
static public void Main ()
{
    int N = 5;
  
    constructmatrix(N);
}
}
  
// This code is contributed by piyush3010 
Output: 
1 2 -2 2 -2 
2 1 -2 2 -2 
2 -2 1 2 -2 
2 -2 2 1 -2 
2 -2 2 -2 1

 

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

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