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 square void findNumbers( int n) { int i = 1; while (i <= n) { // Print ith odd number cout << ((2 * i) - 1) << " " ; i++; } } // Driver code int main() { int n = 3; findNumbers(n); } |
Java
// Java implementation of the approach import 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 = 3 findNumber(n) # This code is contributed by Shrikant13 |
C#
// C# implementation of the approach using 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 square function 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 square function 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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