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Print N numbers such that their product is a Perfect Cube

  • Last Updated : 25 Mar, 2021

Given a number N, the task is to find distinct N numbers such that their product is a perfect cube.
Examples: 
 

Input: N = 3 
Output: 1, 8, 27 
Explanation: 
Product of the output numbers = 1 * 8 * 27 = 216, which is the perfect cube of 6 (63 = 216)
Input: N = 2 
Output: 1 8 
Explanation: 
Product of the output numbers = 1 * 8 = 8, which is the perfect cube of 2 (23 = 8) 
 

 

Approach: The solution is based on the fact that 
 

The product of the first ‘N’ Perfect Cube numbers is always a Perfect Cube.

 
So, the Perfect Cube of first N natural numbers will be printed as the output.
For example: 
 

For N = 1 => [1]
    Product is 1
    and cube root of 1 is also 1

For N = 2 => [1, 8]
    Product is 8
    and cube root of 8 is 2

For N = 3 => [1, 8, 27]
    Product is 216
    and cube root of 216 is 6

and so on

Below is the implementation of the above approach:
 

C++




// C++ program to find N numbers such that
// their product is a perfect cube
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the N numbers such
//that their product is a perfect cube
void findNumbers(int N)
{
    int i = 1;
 
    // Loop to traverse each
//number from 1 to N
    while (i <= N) {
 
// Print the cube of i
//as the ith term of the output
        cout << (i * i * i)
             << " ";
 
        i++;
    }
}
 
// Driver Code
int main()
{
    int N = 4;
 
    // Function Call
    findNumbers(N);
}

Java




// Java program to find N numbers such that
// their product is a perfect cube
import java.util.*;
 
class GFG
{
     
// Function to find the N numbers such
//that their product is a perfect cube
static void findNumbers(int N)
{
    int i = 1;
 
    // Loop to traverse each
    //number from 1 to N
    while (i <= N) {
 
        // Print the cube of i
        //as the ith term of the output
        System.out.print( (i * i * i)
            + " ");
 
        i++;
    }
}
 
// Driver Code
public static void main (String []args)
{
    int N = 4;
 
    // Function Call
    findNumbers(N);
}
}
 
// This code is contributed by chitranayal

Python3




# Python3 program to find N numbers such that
# their product is a perfect cube
 
# Function to find the N numbers such
# that their product is a perfect cube
def findNumbers(N):
    i = 1
 
    # Loop to traverse each
    # number from 1 to N
    while (i <= N):
     
        # Print the cube of i
        # as the ith term of the output
        print((i * i * i), end=" ")
 
        i += 1
 
# Driver Code
if __name__ == '__main__':
    N = 4
 
    # Function Call
    findNumbers(N)
 
# This code is contributed by mohit kumar 29

C#




// C# program to find N numbers such that
// their product is a perfect cube
using System;
 
class GFG
{
     
// Function to find the N numbers such
//that their product is a perfect cube
static void findNumbers(int N)
{
    int i = 1;
 
    // Loop to traverse each
    //number from 1 to N
    while (i <= N) {
 
        // Print the cube of i
        //as the ith term of the output
        Console.Write( (i * i * i)
            + " ");
 
        i++;
    }
}
 
// Driver Code
public static void Main (string []args)
{
    int N = 4;
 
    // Function Call
    findNumbers(N);
}
}
 
// This code is contributed by Yash_R

Javascript




<script>
// JavaScript program to find N numbers such that
// their product is a perfect cube
 
// Function to find the N numbers such
//that their product is a perfect cube
function findNumbers(N)
{
    let i = 1;
 
    // Loop to traverse each
    //number from 1 to N
    while (i <= N) {
 
    // Print the cube of i
    //as the ith term of the output
        document.write((i * i * i)
            + " ");
 
        i++;
    }
}
 
// Driver Code
 
    let N = 4;
 
    // Function Call
    findNumbers(N);
 
// This code is contributed by Manoj.
</script>
Output: 
1 8 27 64

 

Performance Analysis: 
 

  • Time Complexity: As in the above approach, we are finding the Perfect Cube of N numbers, therefore it will take O(N) time.
  • Auxiliary Space Complexity: As in the above approach, there are no extra space used; therefore the Auxiliary Space complexity will be O(1)
     

 


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