# Minimum divisor of a number to make the number perfect cube

Given a positive integer **N**, the task is to find the minimum divisor by which it shall be divided to make it a perfect cube. If N is already a perfect cube, then print 1.**Examples:**

Input: N = 128Output: 2 By Dividing N by 2, we get 64 which is a perfect cube.Input: n = 6Output: 6 By Dividing N by 6, we get 1 which is a perfect cube.Input: n = 64Output: 1

Any number is a perfect cube if all prime factors of it appear in multiples of 3, as you can see in the below figure.

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Therefore, the idea is to find the prime factorization of N and find power of each prime factor. Now, find and multiply all the prime factors whose power is not divisible by 3 as primeFactor*^{power%3}. The resultant of the multiplication is the answer.

Below is the implementation of the above approach:

## C++

`// C++ program to find minimum number which divide n` `// to make it a perfect cube` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Returns the minimum divisor` `int` `findMinNumber(` `int` `n)` `{` ` ` `int` `count = 0, ans = 1;` ` ` `// Since 2 is only even prime, compute its` ` ` `// power seprately.` ` ` `while` `(n % 2 == 0) {` ` ` `count++;` ` ` `n /= 2;` ` ` `}` ` ` `// If count is not divisible by 3,` ` ` `// it must be removed by dividing` ` ` `// n by prime number power.` ` ` `if` `(count % 3 != 0)` ` ` `ans *= ` `pow` `(2, (count % 3));` ` ` `for` `(` `int` `i = 3; i <= ` `sqrt` `(n); i += 2) {` ` ` `count = 0;` ` ` `while` `(n % i == 0) {` ` ` `count++;` ` ` `n /= i;` ` ` `}` ` ` `// If count is not divisible by 3,` ` ` `// it must be removed by dividing` ` ` `// n by prime number power.` ` ` `if` `(count % 3 != 0)` ` ` `ans *= ` `pow` `(i, (count % 3));` ` ` `}` ` ` `// if n is a prime number` ` ` `if` `(n > 2)` ` ` `ans *= n;` ` ` `return` `ans;` `}` `// Driven Program` `int` `main()` `{` ` ` `int` `n = 128;` ` ` `cout << findMinNumber(n) << endl;` ` ` `return` `0;` `}` |

## Java

`// Java program to find minimum number which divide n` `// to make it a perfect cube` `class` `GFG{` ` ` `// Returns the minimum divisor` `static` `int` `findMinNumber(` `int` `n)` `{` ` ` `int` `count = ` `0` `, ans = ` `1` `;` ` ` ` ` `// Since 2 is only even prime, compute its` ` ` `// power seprately.` ` ` `while` `(n % ` `2` `== ` `0` `) {` ` ` `count++;` ` ` `n /= ` `2` `;` ` ` `}` ` ` ` ` `// If count is not divisible by 3,` ` ` `// it must be removed by dividing` ` ` `// n by prime number power.` ` ` `if` `(count % ` `3` `!= ` `0` `)` ` ` `ans *= Math.pow(` `2` `, (count % ` `3` `));` ` ` ` ` `for` `(` `int` `i = ` `3` `; i <= Math.sqrt(n); i += ` `2` `) {` ` ` `count = ` `0` `;` ` ` `while` `(n % i == ` `0` `) {` ` ` `count++;` ` ` `n /= i;` ` ` `}` ` ` ` ` `// If count is not divisible by 3,` ` ` `// it must be removed by dividing` ` ` `// n by prime number power.` ` ` `if` `(count % ` `3` `!= ` `0` `)` ` ` `ans *= Math.pow(i, (count % ` `3` `));` ` ` `}` ` ` ` ` `// if n is a prime number` ` ` `if` `(n > ` `2` `)` ` ` `ans *= n;` ` ` ` ` `return` `ans;` `}` ` ` `// Driver code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `n = ` `128` `;` ` ` `System.out.print(findMinNumber(n) +` `"\n"` `);` `}` `}` `// This code is contributed by Rajput-Ji` |

## Python3

`# Python3 program to find minimum number which divide n` `# to make it a perfect cube` `# Returns the minimum divisor` `def` `findMinNumber(n):` ` ` `count ` `=` `0` `;` ` ` `ans ` `=` `1` `;` ` ` `# Since 2 is only even prime, compute its` ` ` `# power seprately.` ` ` `while` `(n ` `%` `2` `=` `=` `0` `):` ` ` `count` `+` `=` `1` `;` ` ` `n ` `/` `=` `2` `;` ` ` ` ` `# If count is not divisible by 3,` ` ` `# it must be removed by dividing` ` ` `# n by prime number power.` ` ` `if` `(count ` `%` `3` `!` `=` `0` `):` ` ` `ans ` `*` `=` `pow` `(` `2` `, (count ` `%` `3` `));` ` ` `for` `i ` `in` `range` `(` `3` `, ` `int` `(` `pow` `(n, ` `1` `/` `2` `)), ` `2` `):` ` ` `count ` `=` `0` `;` ` ` `while` `(n ` `%` `i ` `=` `=` `0` `):` ` ` `count ` `+` `=` `1` `;` ` ` `n ` `/` `=` `i;` ` ` ` ` `# If count is not divisible by 3,` ` ` `# it must be removed by dividing` ` ` `# n by prime number power.` ` ` `if` `(count ` `%` `3` `!` `=` `0` `):` ` ` `ans ` `*` `=` `pow` `(i, (count ` `%` `3` `));` ` ` ` ` `# if n is a prime number` ` ` `if` `(n > ` `2` `):` ` ` `ans ` `*` `=` `n;` ` ` `return` `ans;` `# Driver code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` `n ` `=` `128` `;` ` ` `print` `(findMinNumber(n));` `# This code is contributed by 29AjayKumar` |

## C#

`// C# program to find minimum number which divide n` `// to make it a perfect cube` `using` `System;` `class` `GFG{` ` ` `// Returns the minimum divisor` `static` `int` `findMinNumber(` `int` `n)` `{` ` ` `int` `count = 0, ans = 1;` ` ` ` ` `// Since 2 is only even prime, compute its` ` ` `// power seprately.` ` ` `while` `(n % 2 == 0) {` ` ` `count++;` ` ` `n /= 2;` ` ` `}` ` ` ` ` `// If count is not divisible by 3,` ` ` `// it must be removed by dividing` ` ` `// n by prime number power.` ` ` `if` `(count % 3 != 0)` ` ` `ans *= (` `int` `)Math.Pow(2, (count % 3));` ` ` ` ` `for` `(` `int` `i = 3; i <= Math.Sqrt(n); i += 2) {` ` ` `count = 0;` ` ` `while` `(n % i == 0) {` ` ` `count++;` ` ` `n /= i;` ` ` `}` ` ` ` ` `// If count is not divisible by 3,` ` ` `// it must be removed by dividing` ` ` `// n by prime number power.` ` ` `if` `(count % 3 != 0)` ` ` `ans *= (` `int` `)Math.Pow(i, (count % 3));` ` ` `}` ` ` ` ` `// if n is a prime number` ` ` `if` `(n > 2)` ` ` `ans *= n;` ` ` ` ` `return` `ans;` `}` ` ` `// Driver code` `public` `static` `void` `Main(String[] args)` `{` ` ` `int` `n = 128;` ` ` `Console.Write(findMinNumber(n) +` `"\n"` `);` `}` `}` `// This code is contributed by 29AjayKumar` |

## Javascript

`<script>` `// Javascript program to find minimum number which divide n` `// to make it a perfect cube` `// Returns the minimum divisor` `function` `findMinNumber(n)` `{` ` ` `var` `count = 0, ans = 1;` ` ` `// Since 2 is only even prime, compute its` ` ` `// power seprately.` ` ` `while` `(n % 2 == 0) {` ` ` `count++;` ` ` `n /= 2;` ` ` `}` ` ` `// If count is not divisible by 3,` ` ` `// it must be removed by dividing` ` ` `// n by prime number power.` ` ` `if` `(count % 3 != 0)` ` ` `ans *= Math.pow(2, (count % 3));` ` ` `for` `(` `var` `i = 3; i <= Math.sqrt(n); i += 2) {` ` ` `count = 0;` ` ` `while` `(n % i == 0) {` ` ` `count++;` ` ` `n /= i;` ` ` `}` ` ` `// If count is not divisible by 3,` ` ` `// it must be removed by dividing` ` ` `// n by prime number power.` ` ` `if` `(count % 3 != 0)` ` ` `ans *= Math.pow(i, (count % 3));` ` ` `}` ` ` `// if n is a prime number` ` ` `if` `(n > 2)` ` ` `ans *= n;` ` ` `return` `ans;` `}` `// Driven Program` `var` `n = 128;` `document.write(findMinNumber(n));` `// This code is contributed by rutvik_56.` `</script>` |

**Output:**

2

**Time Complexity: **O(sqrt(n))

**Auxiliary Space: **O(1)