# 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```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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 conributed by chitranayal `

Output:

```Count of Cubes is 2
```

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 ` `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 `

Output:

```Count of cubes is 2
```

Time Complexity: O(Log b).

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