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 8Input: 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 |
7 0 0 0 7 0 0 0 7
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