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 matrixvoid 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 codeint main(){ int n = 3, k = 7; printMatrix(n, k); return (0);} |
Java
// Java implementation of the approachimport java.util.*;class GFG {// Function to print the required matrixstatic 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 codepublic 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 approachusing System; class GFG {// Function to print the required matrixstatic 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 codepublic 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 matrixfunction 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 codevar 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.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
