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# 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 ``#include ` `// 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 ``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

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## Javascript

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Output:

`6`

Time Complexity: O(N1/6)

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

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