Program to count number of distinct Squares and Cubes upto N

Given a number N, the task is to find No. of perfect Squares and perfect Cubes from 1 to a given integer, N ( Both Inclusive).
Note: Numbers which are both perfect Square and perfect Cube should be counted once.
Examples:

Input: N = 70 
Output: 10 
Explanation: Numbers that are perfect Square or perfect Cube 
1, 4, 8, 9, 16, 25, 27, 36, 49, 64

Input: N = 25 
Output: 6 
Explanation: Numbers that are perfect Square or perfect Cube 
1, 4, 8, 9, 16, 25  

Naive Approach: 

Check all numbers from 1 to N, if it is a square or cube.

Below is the implementation of above approach:  

C++

// C++ implementation of above approach
#include <iostream>
#include <math.h> // For sqrt() and cbrt()
using namespace std;

// Function to check if the
// number is a perfect square
bool isSquare(int num)
{
    int root = sqrt(num);
    return (root * root) == num;
}

// Function to check if the
// number is a perfect cube
bool isCube(int num)
{
    int root = cbrt(num);
    return (root * root * root) == num;
}

// Function to count the number
// of perfect squares and cubes
int countSC(int N)
{
    int count = 0;
    for (int i = 1; i <= N; i++) {

        // If a number is perfect square,
        if (isSquare(i))
            count++;

        // Else if the number is cube or not
        else if (isCube(i))
            count++;
    }

    return count;
}

// Driver code
int main()
{
    int N = 20;

    cout << "Number of squares and cubes is " << countSC(N);
    return 0;
}

Java

// Java implementation of
// above approach
class GFG {

    // Function to check if the
    // number is a perfect square
    static boolean isSquare(int num)
    {
        int root = (int)Math.sqrt(num);
        return (root * root) == num;
    }

    // Function to check if the
    // number is a perfect cube
    static boolean isCube(int num)
    {
        int root = (int)Math.cbrt(num);
        return (root * root * root) == num;
    }

    // Function to count the number
    // of perfect squares and cubes
    static int countSC(int N)
    {
        int count = 0;
        for (int i = 1; i <= N; i++) {

            // If a number is perfect
            // square,
            if (isSquare(i))
                count++;

            // Else if the number is
            // cube or not
            else if (isCube(i))
                count++;
        }

        return count;
    }

    // Driver code
    public static void main(String[] args)
    {
        int N = 20;

        System.out.println("Number of squares "
                           + "and cubes is " + countSC(N));
    }
}

// This code is contributed
// by ChitraNayal

Python3

# Python 3 implementation of
# above approach

# Function to check if the
# number is a perfect square


def isSquare(num):
    root = int(num ** (1 / 2))

    return (root * root) == num

# Function to check if the
# number is a perfect cube


def isCube(num):
    root = int(num ** (1 / 3))
    return (root * root * root) == num

# Function to count the number
# of perfect squares and cubes


def countSC(N):
    count = 0
    for i in range(1, N + 1):

        # If a number is perfect square,
        if isSquare(i):
            count += 1

        # Else if the number is cube
        elif isCube(i):
            count += 1

    return count


# Driver code
if __name__ == "__main__":

    N = 20

    print("Number of squares and cubes is ",
          countSC(N))

# This code is contributed by ANKITRAI1

C#

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

class GFG {

    // Function to check if the
    // number is a perfect square
    static bool isSquare(int num)
    {
        int root = (int)Math.Sqrt(num);
        return (root * root) == num;
    }

    // Function to check if the
    // number is a perfect cube
    static bool isCube(int num)
    {
        int root = (int)Math.Pow(num, (1.0 / 3.0));
        return (root * root * root) == num;
    }

    // Function to count the number
    // of perfect squares and cubes
    static int countSC(int N)
    {
        int count = 0;
        for (int i = 1; i <= N; i++) {

            // If a number is perfect
            // square,
            if (isSquare(i))
                count++;

            // Else if the number is
            // cube or not
            else if (isCube(i))
                count++;
        }

        return count;
    }

    // Driver code
    public static void Main()
    {
        int N = 20;

        Console.Write("Number of squares and "
                      + "cubes is " + countSC(N));
    }
}

// This code is contributed
// by ChitraNayal

PHP

<?php
// PHP implementation of above approach

// Function to check if the
// number is a perfect square
function isSquare($num)
{
    $root = (int)sqrt($num);
    return ($root * $root) == $num;
}

// Function to check if the
// number is a perfect cube
function isCube($num)
{
    $root = (int)pow($num, 1 / 3);
    return ($root * $root *
            $root) == $num;
}

// Function to count the number
// of perfect squares and cubes
function countSC($N)
{
    $count = 0;
    for ($i = 1; $i <= $N; $i++)
    {

        // If a number is
        // perfect square,
        if (isSquare($i))
            $count++;

        // Else if the number
        // is cube or not
        else if (isCube($i))
            $count++;
    }

    return $count;
}

// Driver code
$N = 20;

echo "Number of squares and " .
     "cubes is " . countSC($N);

// This code is contributed
// by ChitraNayal
?>

Javascript

<script>

    // Javascript implementation of above approach

    // Function to check if the
    // number is a perfect square
    function isSquare(num)
    {
        let root = parseInt(Math.sqrt(num), 10);
        return (root * root) == num;
    }

    // Function to check if the
    // number is a perfect cube
    function isCube(num)
    {
        let root = parseInt(Math.cbrt(num), 10);
        return (root * root * root) == num;
    }

    // Function to count the number
    // of perfect squares and cubes
    function countSC(N)
    {
        let count = 0;
        for (let i = 1; i <= N; i++) {

            // If a number is perfect square,
            if (isSquare(i))
                count++;

            // Else if the number is cube or not
            else if (isCube(i))
                count++;
        }

        return count;
    }

    let N = 20;

    document.write("Number of squares and cubes is " + countSC(N));

</script>

Output

Number of squares and cubes is 5

Time Complexity: O(N)
Space Complexity: O(1) 

Efficient Approach: 

• Number of squares from 1 to N is floor(sqrt(N)).
• Number of cubes from 1 to N is floor(cbrt(N)).
• Eliminate the numbers which are both square and cube ( like 1, 64…. ) by subtracting floor(sqrt(cbrt(N))) from it.

Below is the implementation of the above approach:

C++

// C++ implementation of above approach
#include <iostream>
#include <math.h> // For sqrt() and cbrt()
using namespace std;

// Function to count the number
// of perfect squares and cubes
int countSC(int N)
{
    int res = (int)sqrt(N) + (int)cbrt(N)
              - (int)(sqrt(cbrt(N)));

    return res;
}

// Driver code
int main()
{
    int N = 20;

    cout << "Number of squares and cubes is " << countSC(N);
    return 0;
}

Java

// Java implementation of
// above approach
class GFG {

    // Function to count the number
    // of perfect squares and cubes
    static int countSC(int N)
    {
        int res = (int)Math.sqrt(N) + (int)Math.cbrt(N)
                  - (int)(Math.sqrt(Math.cbrt(N)));

        return res;
    }

    // Driver code
    public static void main(String[] args)
    {
        int N = 20;

        System.out.println("Number of squares "
                           + "and cubes is " + countSC(N));
    }
}

// This code is contributed
// by ChitraNayal

Python3

# Python implementation of
# above approach
import math  # for sqrt()

# Function to count the number
# of perfect squares and cubes


def cr(N):
    if pow(round(N ** (1 / 3)), 3) == N:
        return round(N ** (1 / 3))

    return int(N ** (1 / 3))


def countSC(N):
    res = (int(math.sqrt(N)) +
           int(cr(N)) -
           int(math.sqrt(cr(N))))

    return res


# Driver code
N = 20
print("Number of squares and cubes is",
      countSC(N))

# This code is contributed
# by vaibhav29498

C#

// C# implementation of
// above approach
using System;
public class GFG {

    // Function to count the number
    // of perfect squares and cubes
    static int countSC(int N)
    {
        int res = (int)(Math.Sqrt(N)
                        + Math.Ceiling(
                            Math.Pow(N, (double)1 / 3))
                        - (Math.Sqrt(Math.Ceiling(
                            Math.Pow(N, (double)1 / 3)))));

        return res;
    }

    // Driver code
    public static void Main()
    {
        int N = 20;

        Console.Write("Number of squares "
                      + "and cubes is " + countSC(N));
    }
}

/*This code is contributed by 29AjayKumar*/

PHP

<?php
// PHP implementation of above approach

// Function to count the number
// of perfect squares and cubes
function countSC($N)
{
    $res = sqrt($N) + pow($N, 1 / 3) -
                (sqrt(pow($N, 1 / 3)));

    return floor($res);
}

// Driver code
$N = 20;

echo "Number of squares and cubes is " ,
                            countSC($N);

// This code is contributed
// by Shashank
?>

Javascript

<script>

    // Javascript implementation of above approach

    // Function to count the number
    // of perfect squares and cubes
    function countSC(N)
    {
        let res = parseInt(Math.sqrt(N), 10) + parseInt(Math.cbrt(N), 10) - parseInt(Math.sqrt(parseInt(Math.cbrt(N), 10)), 10);

        return res;
    }

    let N = 20;

    document.write("Number of squares and cubes is " + countSC(N));

</script>

Output

Number of squares and cubes is 5

Time Complexity: O(log(N))
Auxiliary Space: O(1)