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Count numbers upto N which are both perfect square and perfect cube
  • Difficulty Level : Medium
  • Last Updated : 26 Mar, 2021

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

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(N1/6)

Auxiliary Space: O(1)

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