Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

Minimum divisor of a number to make the number perfect cube

  • Last Updated : 08 Apr, 2021

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

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

Input : n = 64
Output : 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.  

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.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

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)




My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!