Number of perfect cubes between two given numbers

Given two given numbers a and b where 1<=a<=b, find the number of perfect cubes between a and b (a and b inclusive).
Examples: 

Input :  a = 3, b = 16
Output : 1
The only perfect cube in given range is 8.

Input : a = 7, b = 30
Output : 2
The two cubes in given range are 8, 
and 27

Method 1 : One naive approach is to check all the numbers between a and b (inclusive a and b) and increase count by one whenever we encounter a perfect cube.

Below is the implementation of above idea: 

CPP

// A Simple Method to count cubes between a and b
#include <bits/stdc++.h>
using namespace std;

// Function to count cubes between two numbers
int countCubes(int a, int b)
{
    int cnt = 0; // Initialize result

    // Traverse through all numbers
    for (int i = a; i <= b; i++)

        // Check if current number 'i' is perfect
        // cube
        for (int j = 1; j * j * j <= i; j++)
            if (j * j * j == i)
                cnt++;

    return cnt;
}

// Driver code
int main()
{
    int a = 7, b = 30;
    cout << "Count of Cubes is "
         << countCubes(a, b);
    return 0;
}

Java

// A Simple Method to count cubes between a and b

class GFG{

// Function to count cubes between two numbers
static int countCubes(int a, int b)
{
    int cnt = 0; // Initialize result

    // Traverse through all numbers
    for (int i = a; i <= b; i++)

        // Check if current number 'i' is perfect
        // cube
        for (int j = 1; j * j * j <= i; j++)
            if (j * j * j == i)
                cnt++;

    return cnt;
}

// Driver code
public static void main(String[] args)
{
    int a = 7, b = 30;
    System.out.print("Count of Cubes is "
         + countCubes(a, b));
}
}

// This code is contributed by 29AjayKumar

Python3

# A Simple Method to count cubes between a and b

# Function to count cubes between two numbers
def countCubes(a, b):
    cnt = 0 # Initialize result

    # Traverse through all numbers
    for i in range(a,b+1):

        # Check if current number 'i' is perfect
        # cube
        for j in range(i+1):
            if j*j*j>i:
                break
            if j * j * j == i:
                cnt+=1

    return cnt

# Driver code
if __name__ == '__main__':
    a = 7
    b = 30
    print("Count of Cubes is ",countCubes(a, b))

# This code is contributed by mohit kumar 29

C#

// A Simple Method to count cubes between a and b
using System;

class GFG{

// Function to count cubes between two numbers
static int countCubes(int a, int b)
{
    int cnt = 0; // Initialize result

    // Traverse through all numbers
    for (int i = a; i <= b; i++)

        // Check if current number 'i' is perfect
        // cube
        for (int j = 1; j * j * j <= i; j++)
            if (j * j * j == i)
                cnt++;

    return cnt;
}

// Driver code
public static void Main()
{
    int a = 7, b = 30;
    Console.Write("Count of Cubes is "
         + countCubes(a, b));
}
}

// This code is contributed by chitranayal

Javascript

<script>
// JavaScript program to count cubes between a and b

    // Function to count cubes between two numbers
    function countCubes(a, b)
    {
        let cnt = 0; // Initialize result

        // Traverse through all numbers
        for (let i = a; i <= b; i++)

            // Check if current number 'i' is perfect
            // cube
            for (let j = 1; j * j * j <= i; j++)
                if (j * j * j == i)
                    cnt++;

        return cnt;
    }

    // Driver code
    let a = 7, b = 30;
    document.write("Count of Cubes is " + countCubes(a, b));

// This code is contributed by Surbhi Tyagi

</script>

Output:

Count of Cubes is 2

Time Complexity: O((b – a)^4/3)

Auxiliary Space: O(1)

Method 2 (Efficient): We can simply take cube root of ‘a’ and cube root of ‘b’ and Round cube root of ‘a’ up and cube root of ‘b’ down and count the perfect cubes between them using: 

(floor(cbrt(b)) - ceil(cbrt(a)) + 1)

We take floor of cbrt(b) because we need to consider 
numbers before b.

We take ceil of cbrt(a) because we need to consider 
numbers after a.


For example, let b = 28, a = 7.  floor(cbrt(b)) = 3, 
ceil(cbrt(a)) = 2.  And number of cubes is 3 - 2 + 1
= 2. The two numbers are 8 and 27.

Below is the implementation of above idea :

C++

// An Efficient Method to count cubes between a and b
#include <bits/stdc++.h>
using namespace std;

// Function to count cubes between two numbers
int countCubes(int a, int b)
{
    return (floor(cbrt(b)) - ceil(cbrt(a)) + 1);
}

// Driver code
int main()
{
    int a = 7, b = 28;
    cout << "Count of cubes is "
         << countCubes(a, b);
    return 0;
}

Java

// An Efficient Method to count cubes between a and b
class GFG
{

// Function to count cubes between two numbers
static int countCubes(int a, int b)
{
    return (int) (Math.floor(Math.cbrt(b)) -
                Math.ceil(Math.cbrt(a)) + 1);
}

// Driver code
public static void main(String[] args)
{
    int a = 7, b = 28;
    System.out.print("Count of cubes is "
        + countCubes(a, b));
}
}

// This code is contributed by 29AjayKumar

Python3

# An Efficient Method to count cubes between a and b
from math import *

# Function to count cubes between two numbers
def countCubes(a, b):

    return (floor(b **(1./3.)) - ceil(a **(1./3.)) + 1)

# Driver code
a = 7
b = 28
print("Count of cubes is",countCubes(a, b))

# This code is contributed by shubhamsingh10

C#

// An Efficient Method to count cubes between a and b
// C# implementation of the above approach
using System;

class GFG
{

// Function to count cubes between two numbers
static int countCubes(int a, int b)
{
    return (int) (Math.Floor(Math.Cbrt(b)) -
                Math.Ceiling(Math.Cbrt(a)) + 1);
}

// Driver code
public static void Main(string[] args)
{
    int a = 7, b = 28;
    Console.WriteLine("Count of cubes is " + countCubes(a, b));
}
}

// This code is contributed by Yash_R

Javascript

<script>

// An Efficient Method to count cubes between a and b

// Function to count cubes between two numbers
function countCubes(a, b){

    return (Math.floor(b **(1./3.)) - Math.ceil(a **(1./3.)) + 1)
}

// Driver code
let a = 7;
let b = 28;
document.write("Count of cubes is "+countCubes(a, b))

// This code is contributed
// by pulamolu mohan pavan cse
</script>

Output:

Count of cubes is 2

Time Complexity: O(Log b).

Auxiliary Space: O(1)