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

• Last Updated : 28 Mar, 2022

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 ``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

 ``
Output:
```7 0 0
0 7 0
0 0 7```

Time Complexity: O(n2)

Auxiliary Space: O(1)

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