# Number of times the largest Perfect Cube can be subtracted from N

Given a number N, at every step, subtract the largest perfect cube( ≤ N) from N. Repeat this step while N > 0. The task is to count the number of steps that can be performed.

Examples:

Input: N = 100
Output: 4
First step, 100 – (4 * 4 * 4) = 100 – 64 = 36
Second step, 36 – (3 * 3 * 3) = 36 – 27 = 9
Third step, 9 – (2 * 2 * 2) = 9 – 8 = 1
Fourth step, 1 – (1 * 1 * 1) = 1 – 1 = 0

Input: N = 150
Output: 5
First step, 150 – (5 * 5 * 5) = 150 – 125 = 25
Second step, 25 – (2 * 2 * 2) = 25 – 8 = 17
Third step, 17 – (2 * 2 * 2) = 17 – 8 = 9
Fourth step, 9 – (2 * 2 * 2) = 9 – 8 = 1
Fifth step, 1 – (1 * 1 * 1) = 1 – 1 = 0

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

Approach:

1. Get the number from which the largest perfect cube has to be reduced.
2. Find the cube root of the number and convert the result as an integer. The cube root of the number might contain some fraction part after the decimal, which needs to be avoided.
3. Subtract the cube of the integer found in the previous step. This would remove the largest possible perfect cube from the number in the above step.
```N = N - ((int) ∛N)3
```
4. Repeat the above two steps with the reduced number, till it is greater than 0.
5. Print the number of times a perfect cube has been reduced from N. This is the final result.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to return the count of steps ` `int` `countSteps(``int` `n) ` `{ ` ` `  `    ``// Variable to store the count of steps ` `    ``int` `steps = 0; ` ` `  `    ``// Iterate while N > 0 ` `    ``while` `(n) { ` ` `  `        ``// Get the largest perfect cube ` `        ``// and subtract it from N ` `        ``int` `largest = cbrt(n); ` `        ``n -= (largest * largest * largest); ` ` `  `        ``// Increment steps ` `        ``steps++; ` `    ``} ` ` `  `    ``// Return the required count ` `    ``return` `steps; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 150; ` `    ``cout << countSteps(n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG{ ` `  `  `// Function to return the count of steps ` `static` `int` `countSteps(``int` `n) ` `{ ` `  `  `    ``// Variable to store the count of steps ` `    ``int` `steps = ``0``; ` `  `  `    ``// Iterate while N > 0 ` `    ``while` `(n > ``0``) { ` `  `  `        ``// Get the largest perfect cube ` `        ``// and subtract it from N ` `        ``int` `largest = (``int``) Math.cbrt(n); ` `        ``n -= (largest * largest * largest); ` `  `  `        ``// Increment steps ` `        ``steps++; ` `    ``} ` `  `  `    ``// Return the required count ` `    ``return` `steps; ` `} ` `  `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `n = ``150``; ` `    ``System.out.print(countSteps(n));  ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

## Python3

 `# Python3 implementation of the approach ` `from` `math ``import` `floor ` ` `  `# Function to return the count of steps ` `def` `countSteps(n): ` ` `  `    ``# Variable to store the count of steps ` `    ``steps ``=` `0` ` `  `    ``# Iterate while N > 0 ` `    ``while` `(n): ` ` `  `        ``# Get the largest perfect cube ` `        ``# and subtract it from N ` `        ``largest ``=` `floor(n``*``*``(``1``/``3``)) ` `        ``n ``-``=` `(largest ``*` `largest ``*` `largest) ` ` `  `        ``# Increment steps ` `        ``steps ``+``=` `1` ` `  `    ``# Return the required count ` `    ``return` `steps ` ` `  `# Driver code ` `n ``=` `150` `print``(countSteps(n)) ` ` `  `# This code is contributed by mohit kumar 29 `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG{ ` `   `  `// Function to return the count of steps ` `static` `int` `countSteps(``int` `n) ` `{ ` `   `  `    ``// Variable to store the count of steps ` `    ``int` `steps = 0; ` `   `  `    ``// Iterate while N > 0 ` `    ``while` `(n > 0) { ` `   `  `        ``// Get the largest perfect cube ` `        ``// and subtract it from N ` `        ``int` `largest = (``int``) Math.Pow(n,(``double``)1/3); ` `        ``n -= (largest * largest * largest); ` `   `  `        ``// Increment steps ` `        ``steps++; ` `    ``} ` `   `  `    ``// Return the required count ` `    ``return` `steps; ` `} ` `   `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `n = 150; ` `    ``Console.Write(countSteps(n));  ` `} ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `

Output:

```5
```

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.