# Least number to be added to or subtracted from N to make it a Perfect Cube

• Last Updated : 30 Mar, 2020

Given a number N, Find the minimum number that needs to be added to or subtracted from N, to make it a perfect cube. If the number is to be added, print it with a + sign, else if the number is to be subtracted, print it with a – sign.

Examples:

Input: N = 25
Output: 2
Nearest perfect cube before 25 = 8
Nearest perfect cube after 25 = 27
Therefore 2 needs to be added to 25 to get the closest perfect cube

Input: N = 40
Output: -13
Nearest perfect cube before 40 = 25
Nearest perfect cube after 40 = 64
Therefore 13 needs to be subtracted from 40 to get the closest perfect cube

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

Approach:

1. Get the number.
2. Find the cube root of the number and convert the result as an integer.
3. After converting the double value to integer, this will contain the root of the perfect cube before N, i.e. floor(cube root(N)).
4. Then find the cube of this number, which will be the perfect cube before N.
5. Find the root of the perfect cube after N, i.e. the ceil(cube root(N)).
6. Then find the cube of this number, which will be the perfect cube after N.
7. Check whether the cube of floor value is nearest to N or the ceil value.
8. If the cube of floor value is nearest to N, print the difference with a -sign. Else print the difference between the cube of the ceil value and N with a + sign.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach`` ` `#include ``using` `namespace` `std;`` ` `// Function to return the Least number``int` `nearest(``int` `n)``{`` ` `    ``// Get the perfect cube``    ``// before and after N``    ``int` `prevCube = cbrt(n);``    ``int` `nextCube = prevCube + 1;``    ``prevCube = prevCube * prevCube * prevCube;``    ``nextCube = nextCube * nextCube * nextCube;`` ` `    ``// Check which is nearest to N``    ``int` `ans``        ``= (n - prevCube) < (nextCube - n)``              ``? (prevCube - n)``              ``: (nextCube - n);`` ` `    ``// return the result``    ``return` `ans;``}`` ` `// Driver code``int` `main()``{``    ``int` `n = 25;``    ``cout << nearest(n) << endl;`` ` `    ``n = 27;``    ``cout << nearest(n) << endl;`` ` `    ``n = 40;``    ``cout << nearest(n) << endl;`` ` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach ``class` `GFG {``     ` `    ``// Function to return the Least number ``    ``static` `int` `nearest(``int` `n) ``    ``{ ``     ` `        ``// Get the perfect cube ``        ``// before and after N ``        ``int` `prevCube = (``int``)Math.cbrt(n); ``        ``int` `nextCube = prevCube + ``1``; ``        ``prevCube = prevCube * prevCube * prevCube; ``        ``nextCube = nextCube * nextCube * nextCube; ``     ` `        ``// Check which is nearest to N ``        ``int` `ans = (n - prevCube) < (nextCube - n) ? ``                    ``(prevCube - n) : (nextCube - n); ``     ` `        ``// return the result ``        ``return` `ans; ``    ``} ``     ` `    ``// Driver code ``    ``public` `static` `void` `main (String[] args)``    ``{ ``        ``int` `n = ``25``; ``        ``System.out.println(nearest(n)); ``     ` `        ``n = ``27``; ``        ``System.out.println(nearest(n)) ; ``     ` `        ``n = ``40``; ``        ``System.out.println(nearest(n)) ; ``    ``} ``}`` ` `// This code is contributed by Yash_R`

Output:

```2
0
-13
```

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