# Check whether N can be a Perfect Cube after adding or subtracting K

Given two integers N and K, the task is to check whether N can be made a perfect cube after adding to or subtracting from K to or from it.

Examples:

Input: N = 7, K = 1
Output: Yes
7 + 1 = 8 which is a perfect cube (23 = 8)

Input: N = 5, K = 4
Output: Yes
5 – 4 = 1 which is a perfect cube (13 = 1)

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

Approach: The simplest way to solve this problem is to check whether either (N + K) or (N – K) is a perfect cube or not.

1. Check whether (N + K) is a perfect cube or not
2. If not, then check whether (N – K) is a perfect cube or not.
3. If both are not perfect cube, then print “No”, else print “Yes”.
4. In order to check whether a number is a perfect cube or not, the easiest way is to find the cube of the floor value of cube root of the number, and then check whether this cube is same as the number or not.
```if(N3 == (floor(∛N))3)
Then N is a perfect cube
```

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above approach ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to check if a number is ` `// a perfect Cube or not ` `bool` `isPerfectCube(``int` `x) ` `{ ` `    ``int` `cr = round(cbrt(x)); ` `    ``return` `(cr * cr * cr == x); ` `} ` ` `  `void` `canBePerfectCube(``int` `N, ``int` `K) ` `{ ` `    ``if` `(isPerfectCube(N + K) ` `        ``|| isPerfectCube(N - K)) ` `        ``cout << ``"Yes\n"``; ` `    ``else` `        ``cout << ``"No\n"``; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `N = 7, K = 1; ` `    ``canBePerfectCube(N, K); ` ` `  `    ``N = 5, K = 4; ` `    ``canBePerfectCube(N, K); ` ` `  `    ``N = 7, K = 2; ` `    ``canBePerfectCube(N, K); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the above approach ` `class` `GFG { ` ` `  `    ``// Function to check if a number is ` `    ``// a perfect Cube or not ` `    ``static` `boolean` `isPerfectCube(``int` `x) ` `    ``{ ` `        ``int` `cr = (``int``)Math.cbrt(x); ` `        ``return` `(cr * cr * cr == x); ` `    ``} ` ` `  `    ``static` `void` `canBePerfectCube(``int` `N, ``int` `K) ` `    ``{ ` `        ``if` `(isPerfectCube(N + K) ` `            ``|| isPerfectCube(N - K) == ``true``) ` `            ``System.out.println(``"Yes"``); ` `        ``else` `             ``System.out.println(``"No"``); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main (String[] args) ` `    ``{ ` `        ``int` `N = ``7``; ` `        ``int` `K = ``1``; ` `        ``canBePerfectCube(N, K); ` `     `  `        ``N = ``5``;  ` `        ``K = ``4``; ` `        ``canBePerfectCube(N, K); ` `     `  `        ``N = ``7``; K = ``2``; ` `        ``canBePerfectCube(N, K); ` `     `  `    ``} ` `} ` ` `  `// This code is contributed by Yash_R `

## Python3

 `# Python3 implementation of the above approach ` ` `  `# Function to check if a number is ` `# a perfect Cube or not ` `def` `isPerfectCube(x) : ` `    ``cr ``=` `int``(x ``*``*` `(``1``/``3``)); ` `    ``return` `(cr ``*` `cr ``*` `cr ``=``=` `x); ` ` `  `def` `canBePerfectCube(N, K) : ` `    ``if` `(isPerfectCube(N ``+` `K) ``or` `isPerfectCube(N ``-` `K)) : ` `        ``print``(``"Yes"``); ` `    ``else` `: ` `        ``print``(``"No"``); ` ` `  `# Driver code ` `if` `__name__ ``=``=` `"__main__"` `:  ` ` `  `    ``N ``=` `7``; K ``=` `1``; ` `    ``canBePerfectCube(N, K); ` ` `  `    ``N ``=` `5``; K ``=` `4``; ` `    ``canBePerfectCube(N, K); ` ` `  `    ``N ``=` `7``; K ``=` `2``; ` `    ``canBePerfectCube(N, K); ` ` `  `# This code is contributed by Yash_R `

## C#

 `// C# implementation of the above approach ` `using` `System; ` ` `  `class` `GFG { ` ` `  `    ``// Function to check if a number is ` `    ``// a perfect Cube or not ` `    ``static` `bool` `isPerfectCube(``int` `x) ` `    ``{ ` `        ``int` `cr = (``int``)Math.Cbrt(x); ` `        ``return` `(cr * cr * cr == x); ` `    ``} ` ` `  `    ``static` `void` `canBePerfectCube(``int` `N, ``int` `K) ` `    ``{ ` `        ``if` `(isPerfectCube(N + K) ` `            ``|| isPerfectCube(N - K) == ``true``) ` `            ``Console.WriteLine(``"Yes"``); ` `        ``else` `             ``Console.WriteLine(``"No"``); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main (``string``[] args) ` `    ``{ ` `        ``int` `N = 7; ` `        ``int` `K = 1; ` `        ``canBePerfectCube(N, K); ` `     `  `        ``N = 5;  ` `        ``K = 4; ` `        ``canBePerfectCube(N, K); ` `     `  `        ``N = 7; K = 2; ` `        ``canBePerfectCube(N, K); ` `     `  `    ``} ` `} ` ` `  `// This code is contributed by AnkitRai01 `

Output:

```Yes
Yes
No
```

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