Given an integer n, the task is to print n numbers such that their sum is a perfect square.
Examples:
Input: n = 3
Output: 1 3 5
1 + 3 + 5 = 9 = 32Input: n = 4
Output: 1 3 5 7
1 + 3 + 5 + 7 = 16 = 42
Approach: The sum of first n odd numbers is always a perfect square. So, we will print the first n odd numbers as the output.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to print n numbers such that// their sum is a perfect squarevoid findNumbers(int n){ int i = 1; while (i <= n) { // Print ith odd number cout << ((2 * i) - 1) << " "; i++; }}// Driver codeint main(){ int n = 3; findNumbers(n);} |
Java
// Java implementation of the approachimport java.util.*;import java.io.*;class GFG { // Function to print n numbers such that // their sum is a perfect square static void findNumbers(int n) { int i = 1; while (i <= n) { // Print ith odd number System.out.print(((2 * i) - 1) + " "); i++; } } // Driver code public static void main(String args[]) { int n = 3; findNumbers(n); }} |
Python3
# Python3 implementation of the approach # Function to print n numbers such that # their sum is a perfect square def findNumber(n): i = 1 while i <= n: # Print ith odd number print((2 * i) - 1, end = " ") i += 1# Driver code n = 3findNumber(n)# This code is contributed by Shrikant13 |
C#
// C# implementation of the approachusing System;public class GFG { // Function to print n numbers such that // their sum is a perfect square public static void findNumbers(int n) { int i = 1; while (i <= n) { // Print ith odd number Console.Write(((2 * i) - 1) + " "); i++; } } // Driver code public static void Main(string[] args) { int n = 3; findNumbers(n); }}// This code is contributed by Shrikant13 |
PHP
<?php// PHP implementation of the approach// Function to print n numbers such that// their sum is a perfect squarefunction findNumbers($n){ $i = 1; while ($i <= $n) { // Print ith odd number echo ((2 * $i) - 1) . " "; $i++; }}// Driver code$n = 3;findNumbers($n);// This code contributed by PrinciRaj1992?> |
Javascript
<script>// JavaScript implementation of the approach// Function to print n numbers such that// their sum is a perfect squarefunction findNumbers(n){ var i = 1; while (i <= n) { // Print ith odd number document.write(((2 * i) - 1)+" ") ; i++; }}var n = 3; findNumbers(n);</script> |
Output
1 3 5
Time Complexity: O(N), as we are using a loop to traverse N times so it will cost us O(N) time.
Auxiliary Space: O(1), as we are not using any extra space.
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