Given N integers with unknown values (ai > 0) having product P. The task is to find the maximum possible greatest common divisor of these N integers.
Examples:
Input : N = 3, P = 24 Output : 2 The integers will have maximum GCD of 2 when a1 = 2, a2 = 2, a3 = 6. Input : N = 2, P = 1 Output : 1 Only possibility is a1 = 1 and a2 = 1.
Approach:
- First find all the prime factors of product P and store it in a Hashmap.
- The N integers will have maximum GCD when a prime factor will be common in all the integers.
- So if P = p1k1 * p2k2 * p3k3 …. where p1, p2 … are prime numbers then, maximum GCD which can be obtained will be ans = p1k1 / N * p2k2 / N * p3k3 / N ….
Below is the implementation of the above approach:
C++
// C++ implementation of above approach#include <bits/stdc++.h>using namespace std;// Function to find maximum GCD// of N integers with product Pint maxGCD(int N, int P){ int ans = 1; // map to store prime factors of P unordered_map<int, int> prime_factors; // prime factorization of P for (int i = 2; i * i <= P; i++) { while (P % i == 0) { prime_factors[i]++; P /= i; } } if (P != 1) prime_factors[P]++; // traverse all prime factors and // multiply its 1/N power to the result for (auto v : prime_factors) ans *= pow(v.first, v.second / N); return ans;}// Driver codeint main(){ int N = 3, P = 24; cout << maxGCD(N, P); return 0;} |
Java
// Java implementation of above approach import java.util.*;class Solution{// Function to find maximum GCD // of N integers with product P static int maxGCD(int N, int P) { int ans = 1; // map to store prime factors of P Map<Integer, Integer> prime_factors = new HashMap< Integer,Integer>(); // prime factorization of P for (int i = 2; i * i <= P; i++) { while (P % i == 0) { if(prime_factors.get(i)==null) prime_factors.put(i,1); else prime_factors.put(i,(prime_factors.get(i)+1)); P /= i; } } if (P != 1) if(prime_factors.get(P)==null) prime_factors.put(P,1); else prime_factors.put(P,(prime_factors.get(P)+1)); // traverse all prime factors and // multiply its 1/N power to the result Set< Map.Entry< Integer,Integer> > st = prime_factors.entrySet(); for (Map.Entry< Integer,Integer> me:st) { ans *= Math.pow(me.getKey(),me.getValue() / N); } return ans; } // Driver code public static void main(String args[]){ int N = 3, P = 24; System.out.println( maxGCD(N, P)); } }//contributed by Arnab Kundu |
Python3
# Python3 implementation of # above approach from math import sqrt# Function to find maximum GCD # of N integers with product P def maxGCD(N, P): ans = 1 # map to store prime factors of P prime_factors = {} # prime factorization of P for i in range(2, int(sqrt(P) + 1)) : while (P % i == 0) : if i not in prime_factors : prime_factors[i] = 0 prime_factors[i] += 1 P //= i if (P != 1) : prime_factors[P] += 1 # traverse all prime factors and # multiply its 1/N power to the result for key, value in prime_factors.items() : ans *= pow(key, value // N) return ans# Driver code if __name__ == "__main__" : N, P = 3, 24 print(maxGCD(N, P))# This code is contributed by Ryuga |
C#
// C# implementation of above approach using System;using System.Collections.Generic;class GFG{ // Function to find maximum GCD // of N integers with product P static int maxGCD(int N, int P) { int ans = 1; // map to store prime factors of P Dictionary<int, int> prime_factors = new Dictionary< int,int>(); // prime factorization of P for (int i = 2; i * i <= P; i++) { while (P % i == 0) { if(!prime_factors.ContainsKey(i)) prime_factors.Add(i, 1); else prime_factors[i] = prime_factors[i] + 1; P /= i; } } if (P != 1) if(!prime_factors.ContainsKey(P)) prime_factors.Add(P, 1); else prime_factors[P] = prime_factors[P] + 1; // traverse all prime factors and // multiply its 1/N power to the result foreach(KeyValuePair<int, int> me in prime_factors) { ans *= (int)Math.Pow(me.Key,me.Value / N); } return ans; } // Driver code public static void Main(String []args){ int N = 3, P = 24; Console.WriteLine( maxGCD(N, P)); } }// This code is contributed by PrinciRaj1992 |
Javascript
<script>// Javascript implementation of above approach// Function to find maximum GCD// of N integers with product Pfunction maxGCD(N, P) { let ans = 1; // map to store prime factors of P let prime_factors = new Map(); // prime factorization of P for (let i = 2; i * i <= P; i++) { while (P % i == 0) { if (prime_factors.get(i) == null) prime_factors.set(i, 1); else prime_factors.set(i, (prime_factors.get(i) + 1)); P = Math.floor(P / i); } } if (P != 1) prime_factors[P]++; // traverse all prime factors and // multiply its 1/N power to the result console.log(prime_factors) for (let v of prime_factors) { console.log(v) ans *= Math.pow(v[0], Math.floor(v[1] / N)); } return ans;}// Driver codelet N = 3, P = 24;document.write(maxGCD(N, P));// This code is contributed by gfgking</script> |
Output
2
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