Print N numbers such that their product is a Perfect Cube

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++

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// 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);
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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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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