Smallest and Largest N-digit perfect cubes

Given an integer N, the task is to find the smallest and the largest N digit numbers which are also perfect cubes.
Examples:

Input: N = 2 
Output: 27 64 
27 and 64 are the smallest and the largest 2-digit numbers which are also perfect cubes.
Input: N = 3 
Output: 125 729 

Approach: For increasing values of N starting from N = 1, the series will go on like 8, 64, 729, 9261, ….. for the largest N-digit perfect cube whose N^th term will be pow(ceil(cbrt(pow(10, (n))))-1, 3). 
And 1, 27, 125, 1000, ….. for the smallest N-digit perfect cube whose N^th term will be pow(ceil(cbrt(pow(10, (n – 1)))), 3).
Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to print the largest and
// the smallest n-digit perfect cube
void nDigitPerfectCubes(int n)
{

    // Smallest n-digit perfect cube
    cout << pow(ceil(cbrt(pow(10, (n - 1)))), 3) << " ";

    // Largest n-digit perfect cube
    cout << (int)pow(ceil(cbrt(pow(10, (n)))) - 1, 3);
}

// Driver code
int main()
{
    int n = 3;
    nDigitPerfectCubes(n);

    return 0;
}

Java

// Java implementation of the approach
class GFG {

    // Function to print the largest and
    // the smallest n-digit perfect cube
    static void nDigitPerfectCubes(int n)
    {

        // Smallest n-digit perfect cube
        int smallest = (int)Math.pow(Math.ceil(Math.cbrt(Math.pow(10, (n - 1)))), 3);
        System.out.print(smallest + " ");

        int largest = (int)Math.pow(Math.ceil(Math.cbrt(Math.pow(10, (n)))) - 1, 3);
        System.out.print(largest);
    }

    // Driver code
    public static void main(String args[])
    {
        int n = 3;
        nDigitPerfectCubes(n);
    }
}

Python3

# Python3 implementation of the approach
from math import ceil

# Function to print the largest and
# the smallest n-digit perfect cube
def nDigitPerfectCubes(n):

    # Smallest n-digit perfect cube
    print(pow(ceil((pow(10, (n - 1))) **
                       (1 / 3)), 3), end = " ")

    # Largest n-digit perfect cube
    print(pow(ceil((pow(10, (n))) **
                       (1 / 3)) - 1, 3))

# Driver code
if __name__ == "__main__":

    n = 3
    nDigitPerfectCubes(n)

# This code is contributed by Rituraj Jain

C#

// C# implementation of the approach
using System;

class GFG
{

    // Function to print the largest and
    // the smallest n-digit perfect cube
    static void nDigitPerfectCubes(int n)
    {

        // Smallest n-digit perfect cube
        int smallest = (int)Math.Pow(Math.Ceiling(MathF.Cbrt((float)Math.Pow(10, (n - 1)))), 3);
        Console.Write(smallest + " ");

        int largest = (int)Math.Pow(Math.Ceiling(MathF.Cbrt((float)Math.Pow(10, (n)))) - 1, 3);
        Console.Write(largest);
    }

    // Driver code
    static void Main()
    {
        int n = 3;
        nDigitPerfectCubes(n);
    }
}

// This code is contributed by mits

PHP

<?php
// PHP implementation of the approach

// Function to print the largest and
// the smallest n-digit perfect cube
function nDigitPerfectCubes($n)
{

    // Smallest n-digit perfect cube
    print(pow(ceil(pow(pow(10, ($n - 1)),1/3)), 3)." ");

    // Largest n-digit perfect cube
    print((int)pow(ceil(pow(pow(10, ($n)),1/3)) - 1, 3));
}

// Driver code
$n = 3;
nDigitPerfectCubes($n);

// This code is contributed by mits
?>

Javascript

<script>

// javascript implementation of the approach

// Function to print the largest and
// the smallest n-digit perfect cube
function nDigitPerfectCubes( n)
{

    // Smallest n-digit perfect cube
     document.write( Math.pow(Math.ceil(Math.cbrt(Math.pow(10, (n - 1)))), 3) + " ");

    // Largest n-digit perfect cube
     document.write( Math.pow(Math.ceil(Math.cbrt(Math.pow(10, (n)))) - 1, 3));
}

// Driver code

    let n = 3;
    nDigitPerfectCubes(n);

    // This code contributed by aashish1995

</script>

Output:

125 729

Time Complexity: O(log n)

Auxiliary Space: O(1)