Skip to content
Related Articles

Related Articles

Improve Article
Maximum GCD of N integers with given product
  • Last Updated : 20 May, 2021

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 P
int 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 code
int 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 P
function 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 code
 
let N = 3, P = 24;
 
document.write(maxGCD(N, P));
 
// This code is contributed by gfgking
</script>
Output: 
2

 

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 industry experts, please refer Geeks Classes Live 




My Personal Notes arrow_drop_up
Recommended Articles
Page :