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.