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++

// 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). 
  

Approach#2: Using math

This approach iterates through numbers starting from 1 and checks if their cube root is an integer. If it is, then the number is a perfect cube and is printed. The loop continues until n such perfect cubes have been printed.

Algorithm

1. Initialize the count to 0 and i to 1.
2. While count is less than n, do the following:
a. Check if the cube root of i is an integer by comparing the float value of the cube root with its integer value.
b. If the cube root is an integer, print the number i and increment the count.
c. Increment i by 1 after each iteration of the loop.
3. Print a newline character after all the perfect cube numbers have been printed.

C++

#include <iostream>
#include <cmath>

using namespace std;

void print_cube_numbers(int n) {
    int count = 0;
    int i = 1;
    while (count < n) {
        int cube_root = round(cbrt(i));
        if (cube_root * cube_root * cube_root == i) {
            cout << i << " ";
            count++;
        }
        i++;
    }
    cout << endl;
}

int main() {
    print_cube_numbers(3);
    print_cube_numbers(2);
    return 0;
}

Java

import java.util.*;

public class GFG {

    // Function to print the first 'n' cube numbers
    // where cube_root * cube_root * cube_root = i
    public static void printCubeNumbers(int n) {
        int count = 0;
        int i = 1;
        while (count < n) {
            // Calculate the cube root of 'i' using cbrt() method
           // from Math class
            int cubeRoot = (int) Math.round(Math.cbrt(i));

            // Check if cube_root * cube_root * cube_root == i
            if (cubeRoot * cubeRoot * cubeRoot == i) {
                System.out.print(i + " ");
                count++;
            }
            i++;
        }
        System.out.println();
    }

    public static void main(String[] args) {
        printCubeNumbers(3); // Print the first 3 cube numbers
        printCubeNumbers(2); // Print the first 2 cube numbers
    }
}

Python3

import math

def print_cube_numbers(n):
    count = 0
    i = 1
    while count < n:
        if math.pow(i, 1/3) == int(math.pow(i, 1/3)):
            print(i, end=" ")
            count += 1
        i += 1
    print()


print_cube_numbers(3)
print_cube_numbers(2)

C#

using System;

class Program
{
    // Function to print the first 'n' cube numbers
    // where cube_root * cube_root * cube_root = i
    static void PrintCubeNumbers(int n)
    {
        int count = 0;
        int i = 1;
        while (count < n)
        {
            // Calculate the cube root of 'i' using cbrt() method
            // from Math class
            int cubeRoot = (int)Math.Round(Math.Cbrt(i));

            // Check if cube_root * cube_root * cube_root == i
            if (cubeRoot * cubeRoot * cubeRoot == i)
            {
                Console.Write(i + " ");
                count++;
            }
            i++;
        }
        Console.WriteLine();
    }

    static void Main()
    {
        PrintCubeNumbers(3);
        PrintCubeNumbers(2);
    }
}

Javascript

// Function to print the first 'n' cube numbers
// where cube_root * cube_root * cube_root = i
function printCubeNumbers(n) {
    let count = 0;
    let i = 1;
    while (count < n) {
        // Calculate the cube root of 'i'
        let cubeRoot = Math.round(Math.cbrt(i));

        // Check if cube_root * cube_root * cube_root == i
        if (Math.pow(cubeRoot, 3) === i) {
            process.stdout.write(i + " ");
            count++;
        }
        i++;
    }
    console.log();
}

// Entry point
function main() {
    printCubeNumbers(3); // Print the first 3 cube numbers
    printCubeNumbers(2); // Print the first 2 cube numbers
}

main();

Output

1 8 27 
1 8 

Time Complexity: O(N^2/3) because it iterates through numbers from 1 to N and checks the cube root of each number, which takes O(N^1/3) time. Therefore, the overall time complexity is O(N^2/3).

Auxiliary Space:  O(1) because it uses constant space to store the variables count and i, and no additional data structures are used.