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.
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 divisorint findMinNumber(int n){ int count = 0, ans = 1; // Since 2 is only even prime, compute its // power separately. 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 Programint 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 cubeimport java.io.*;public class GFG{ // Returns the minimum divisorstatic int findMinNumber(int n){ int count = 0, ans = 1; // Since 2 is only even prime, compute its // power separately. 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 codepublic 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 divisordef findMinNumber(n): count = 0; ans = 1; # Since 2 is only even prime, compute its # power separately. 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 codeif __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 cubeusing System;class GFG{ // Returns the minimum divisorstatic int findMinNumber(int n){ int count = 0, ans = 1; // Since 2 is only even prime, compute its // power separately. 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 codepublic 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 divisorfunction findMinNumber(n){ var count = 0, ans = 1; // Since 2 is only even prime, compute its // power separately. 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 Programvar 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)
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