Find a Square Matrix such that sum of elements in every row and column is K

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Given two integers N and K, the task is to find an N x N square matrix such that sum of every row and column should be equal to K. Note that there can be multiple such matrices possible. Print any one of them.
Examples:

Input: N = 3, K = 15
Output:
2 7 6
9 5 1
4 3 8
Input: N = 3, K = 7
Output:
7 0 0
0 7 0
0 0 7

Approach: An N x N matrix such that each left diagonal element is equal to K and rest elements are 0 will satisfy the given condition. In this way, the sum of the elements of the each row and column will be equal to K.
Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to print the
// required matrix
void printMatrix(int n, int k)
{
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
 
            // Print k for the left
            // diagonal elements
            if (i == j)
                cout << k << " ";
 
            // Print 0 for the rest
            else
                cout << "0 ";
        }
        cout << "\n";
    }
}
 
// Driver code
int main()
{
    int n = 3, k = 7;
 
    printMatrix(n, k);
 
    return (0);
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG 
{
 
// Function to print the required matrix
static void printMatrix(int n, int k)
{
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < n; j++) 
        {
 
            // Print k for the left
            // diagonal elements
            if (i == j)
                System.out.print(k + " ");
 
            // Print 0 for the rest
            else
                System.out.print("0 ");
        }
        System.out.print("\n");
    }
}
 
// Driver code
public static void main(String[] args)
{
    int n = 3, k = 7;
 
    printMatrix(n, k);
}
} 
 
// This code is contributed by Princi Singh

Python3

# Python3 implementation of the approach 
 
# Function to print the 
# required matrix 
def printMatrix(n, k) :
 
    for i in range(n) :
        for j in range(n) : 
 
            # Print k for the left 
            # diagonal elements 
            if (i == j) :
                print(k, end = " "); 
 
            # Print 0 for the rest 
            else:
                print("0", end = " "); 
                 
        print(); 
 
# Driver code 
if __name__ == "__main__" : 
 
    n = 3; k = 7; 
 
    printMatrix(n, k); 
 
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
     
class GFG 
{
 
// Function to print the required matrix
static void printMatrix(int n, int k)
{
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < n; j++) 
        {
 
            // Print k for the left
            // diagonal elements
            if (i == j)
                Console.Write(k + " ");
 
            // Print 0 for the rest
            else
                Console.Write("0 ");
        }
        Console.Write("\n");
    }
}
 
// Driver code
public static void Main(String[] args)
{
    int n = 3, k = 7;
 
    printMatrix(n, k);
}
}
 
// This code is contributed by PrinciRaj1992

Javascript

<script>
 
// javascript implementation of the approach
 
// Function to print the required matrix
function printMatrix(n , k)
{
    for (i = 0; i < n; i++)
    {
        for (j = 0; j < n; j++) 
        {
 
            // Print k for the left
            // diagonal elements
            if (i == j)
                document.write(k + " ");
 
            // Print 0 for the rest
            else
                document.write("0 ");
        }
        document.write("</br>");
    }
}
 
// Driver code
var n = 3, k = 7;
 
printMatrix(n, k);
 
// This code is contributed by 29AjayKumar 
 
</script>

Output:

7 0 0 
0 7 0 
0 0 7

Time Complexity: O(N*N), as we are using nested loops to traverse N*N times.

Auxiliary Space: O(1), as we are not using any extra space.

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