Count numbers upto N which are both perfect square and perfect cube

Given a number N. The task is to count total numbers under N which are both perfect square and cube of some integers.
Examples:

Input: N = 100
Output: 2
They are 1 and 64.

Input: N = 100000
Output: 6

Naive Approach: The idea is to use the power, square_root and cube_root functions from the math library.

C++

// C++ program for the above approach
#include <cmath>
#include <iostream>

// Function to count the perfect square
// and cubes
int countPerfectSquaresCubes(int N)
{
    int count = 0;

    // Iterate over the range [1, N]
    for (int i = 1; i <= N; i++) {
        if (std::pow((int)std::sqrt(i), 2) == i
            && std::pow((int)std::cbrt(i), 3) == i) {
            count++;
        }
    }

    // Return the resultant count
    return count;
}

// Driver Code
int main()
{
    int count = countPerfectSquaresCubes(100000);
    std::cout << count << std::endl;
    return 0;
}

// This code is contributed by aeroabrar_31

Java

// Java program for the above approach

import java.util.*;
class GFG {

    // Function to count the perfect square
    // and cubes
    public static int countPerfectSquaresCubes(int N)
    {
        int count = 0;
        for (int i = 1; i <= N; i++) {
            if (Math.pow((int)Math.sqrt(i), 2) == i
                && Math.pow((int)Math.cbrt(i), 3) == i) {
                count++;
            }
        }

        // Return the resultant count
        return count;
    }

    // Driver Code
    public static void main(String[] args)
    {
        int count = countPerfectSquaresCubes(100000);
        System.out.println(count);
    }
}

C#

// C# program for the above approach
using System;

public class GFG {

    // Function to count the perfect square
    // and cubes
    public static int countPerfectSquaresCubes(int N)
    {
        int count = 0;
        for (int i = 1; i <= N; i++) {
            if (Math.Pow((int)Math.Sqrt(i), 2) == i
                && Math.Pow((int)Math.Cbrt(i), 3) == i) {
                count++;
            }
        }

        // Return the resultant count
        return count;
    }

    // Driver Code
    public static void Main()
    {
        int count = countPerfectSquaresCubes(100000);
        Console.WriteLine(count);
    }
}
// This code is contributed by aeroabrar_31

Output

6

Time Complexity: O(N*(logN))

Space Complexity: O(1)

Method 2 : Optimal

Approach: For a given positive number N to be a perfect square, it must satisfy P² = N Similarly, Q³ = N for a perfect cube where P and Q are some positive integers. 
N = P² = Q³
Thus, if N is a 6th power, then this would certainly work. Say N = A⁶ which can be written as (A³)² or (A²)³.
So, pick 6th power of every positive integers which are less than equal to N.
Below is the implementation of the above approach:  

C++

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

// Function to return required count
int SquareCube(long long int N)
{

    int cnt = 0, i = 1;

    while (int(pow(i, 6)) <= N) {
        ++cnt;
        ++i;
    }

    return cnt;
}

int main()
{
    long long int N = 100000;

    // function call to print required answer
    cout << SquareCube(N);
    return 0;
}

Java

// Java implementation of the above approach

public class GFG{

    // Function to return required count
    static int SquareCube(long N)
    {

        int cnt = 0, i = 1;

        while ((int)(Math.pow(i, 6)) <= N) {
            ++cnt;
            ++i;
        }

        return cnt;
    }

    public static void main(String []args)
    {
        long N = 100000;

        // function call to print required answer
        System.out.println(SquareCube(N)) ;
    }
    // This code is contributed by Ryuga
}

Python3

# Python3 implementation of the
# above approach

# Function to return required count
def SquareCube( N):

    cnt, i = 0, 1

    while (i ** 6 <= N):
        cnt += 1
        i += 1

    return cnt

# Driver code
N = 100000

# function call to print required
# answer
print(SquareCube(N))

# This code is contributed
# by saurabh_shukla

C#

// C# implementation of the above approach
using System;

public class GFG{

    // Function to return required count
    static int SquareCube(long N)
    {

        int cnt = 0, i = 1;

        while ((int)(Math.Pow(i, 6)) <= N) {
            ++cnt;
            ++i;
        }

        return cnt;
    }

    public static void Main()
    {
        long N = 100000;

        // function call to print required answer
        Console.WriteLine(SquareCube(N)) ;
    }
}

/*This code is contributed by 29AjayKumar*/

PHP

<?php
// PHP implementation of the
// above approach

// Function to return required count
function SquareCube($N)
{
    $cnt = 0;
    $i = 1;

    while ((pow($i, 6)) <= $N)
    {
        ++$cnt;
        ++$i;
    }

    return $cnt;
}

// Driver Code
$N = 100000;

// function call to print required answer
echo SquareCube($N);

// This code is contributed by ita_c
?>

Javascript

<script>
// JavaScript implementation of the above approach

// Function to return required count
function SquareCube(N)
{

    let cnt = 0, i = 1;
    while (Math.floor(Math.pow(i, 6)) <= N)
    {
        ++cnt;
        ++i;
    }
    return cnt;
}

    let N = 100000;

    // function call to print required answer
    document.write(SquareCube(N));

// This code is contributed by Surbhi Tyagi.
</script>

Output:

6

Time Complexity: O(N^1/6)

Auxiliary Space: O(1), since no extra space has been taken.