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Common prime factors of two numbers

  • Difficulty Level : Medium
  • Last Updated : 06 Jun, 2021

Given two integer A   and B   , the task is to find the common prime divisors of these numbers.
Examples: 
 

Input: A = 6, B = 12 
Output: 2 3 
2 and 3 are the only common prime divisors of 6 and 12
Input: A = 4, B = 8 
Output:
 

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Naive Approach: Iterate from 1 to min(A, B) and check whether i is prime and a factor of both A and B, if yes then display the number.
Efficient Approach is to do following: 
 

  1. Find Greatest Common Divisor (gcd) of the given numbers.
  2. Find prime factors of the GCD.

Efficient Approach for multiple queries: The above solution can be further optimized if there are multiple queries for common factors. The idea is based on Prime Factorization using Sieve O(log n) for multiple queries.
Below is the implementation of the above approach: 
 

C++




// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
 
#define MAXN   100001
 
bool prime[MAXN];
void SieveOfEratosthenes()
{
 
    // Create a boolean array "prime[0..n]" and initialize
    // all entries it as true. A value in prime[i] will
    // finally be false if i is Not a prime, else true.
 
    memset(prime, true, sizeof(prime));
 
    // 0 and 1 are not prime numbers
    prime[0] = false;
    prime[1] = false;
 
    for (int p = 2; p * p <= MAXN; p++) {
 
        // If prime[p] is not changed, then it is a prime
        if (prime[p] == true) {
 
            // Update all multiples of p as non-prime
            for (int i = p * p; i <= MAXN; i += p)
                prime[i] = false;
        }
    }
}
 
// Find the common prime divisors
void common_prime(int a, int b)
{
 
    // Get the GCD of the given numbers
    int gcd = __gcd(a, b);
 
    // Find the prime divisors of the gcd
    for (int i = 2; i <= (gcd); i++) {
 
        // If i is prime and a divisor of gcd
        if (prime[i] && gcd % i == 0) {
            cout << i << " ";
        }
    }
}
 
// Driver code
int main()
{
    // Create the Sieve
    SieveOfEratosthenes();
 
    int a = 6, b = 12;
 
    common_prime(a, b);
    return 0;
}

Java




//Java implementation of above approach
 
class GFG {
 
static final int MAXN = 100001;
static boolean prime[] = new boolean[MAXN];
 
static void SieveOfEratosthenes()
{
 
    // Create a boolean array "prime[0..n]" and initialize
    // all entries it as true. A value in prime[i] will
    // finally be false if i is Not a prime, else true.
 
        for(int i = 0;i<prime.length;i++)
            prime[i]=true;
 
    // 0 and 1 are not prime numbers
    prime[0] = false;
    prime[1] = false;
 
    for (int p = 2; p * p < MAXN; p++) {
 
        // If prime[p] is not changed, then it is a prime
        if (prime[p] == true) {
 
            // Update all multiples of p as non-prime
            for (int i = p * p; i < MAXN; i += p)
                prime[i] = false;
        }
    }
}
 
// Find the common prime divisors
static void common_prime(int a, int b)
{
 
    // Get the GCD of the given numbers
    int gcd = (int) __gcd(a, b);
 
    // Find the prime divisors of the gcd
    for (int i = 2; i <= (gcd); i++) {
 
        // If i is prime and a divisor of gcd
        if (prime[i] && gcd % i == 0) {
            System.out.print(i + " ");
        }
    }
}
static long __gcd(long a, long b) 
    if (a == 0
        return b; 
    return __gcd(b % a, a); 
}
// Driver code
    public static void main(String[] args) {
        // Create the Sieve
    SieveOfEratosthenes();
 
    int a = 6, b = 12;
 
    common_prime(a, b);
    }
}
 
/*This code is contributed by 29AjayKumar*/

Python3




# Python implementation of above approach
from math import gcd, sqrt
 
# Create a boolean array "prime[0..n]"
# and initialize all entries it as true.
# A value in prime[i] will finally be
# false if i is Not a prime, else true.
prime = [True] * 100001
 
def SieveOfEratosthenes() :
     
    # 0 and 1 are not prime numbers
    prime[0] = False
    prime[1] = False
     
    for p in range(2, int(sqrt(100001)) + 1) :
 
        # If prime[p] is not changed,
        # then it is a prime
        if prime[p] == True :
 
            # Update all multiples of
            # p as non-prime
            for i in range(p**2, 100001, p) :
                prime[i] = False
     
# Find the common prime divisors
def common_prime(a, b) :
 
    # Get the GCD of the given numbers
    __gcd = gcd(a, b)
 
    # Find the prime divisors of the gcd
    for i in range(2, __gcd + 1) :
  
        # If i is prime and a divisor of gcd
        if prime[i] and __gcd % i == 0 :
            print(i, end = " ")
 
# Driver code
if __name__ == "__main__" :
 
    # Create the Sieve
    SieveOfEratosthenes()
    a, b = 6, 12
     
    common_prime(a, b)
     
# This code is contributed by ANKITRAI1

C#




//C# implementation of above approach
using System;
public class GFG {
  
    static bool []prime = new bool[100001];
    static void SieveOfEratosthenes()
    {
 
        // Create a boolean array "prime[0..n]" and initialize
        // all entries it as true. A value in prime[i] will
        // finally be false if i is Not a prime, else true.
 
            for(int i = 0;i<prime.Length;i++)
                prime[i]=true;
 
        // 0 and 1 are not prime numbers
        prime[0] = false;
        prime[1] = false;
 
        for (int p = 2; p * p < 100001; p++) {
 
            // If prime[p] is not changed, then it is a prime
            if (prime[p] == true) {
 
                // Update all multiples of p as non-prime
                for (int i = p * p; i < 100001; i += p)
                    prime[i] = false;
            }
        }
    }
 
    // Find the common prime divisors
    static void common_prime(int a, int b)
    {
 
        // Get the GCD of the given numbers
        int gcd = (int) __gcd(a, b);
 
        // Find the prime divisors of the gcd
        for (int i = 2; i <= (gcd); i++) {
 
            // If i is prime and a divisor of gcd
            if (prime[i] && gcd % i == 0) {
                Console.Write(i + " ");
            }
        }
    }
    static long __gcd(long a, long b) 
    
        if (a == 0) 
            return b; 
        return __gcd(b % a, a); 
    }
    // Driver code
    public static void Main() {
        // Create the Sieve
    SieveOfEratosthenes();
  
    int a = 6, b = 12;
  
    common_prime(a, b);
    }
}
  
/*This code is contributed by 29AjayKumar*/

Javascript




<script>
// Javascript program to implement the above approach
 
MAXN = parseInt(100001);
prime = new Array(MAXN);
 
function __gcd(a, b) 
    if (a == 0) 
        return b; 
    return __gcd(b % a, a); 
}
function SieveOfEratosthenes()
{
    // Create a boolean array "prime[0..n]" and initialize
    // all entries it as true. A value in prime[i] will
    // finally be false if i is Not a prime, else true.
 
    prime.fill(true);
 
    // 0 and 1 are not prime numbers
    prime[0] = false;
    prime[1] = false;
 
    for (var p = 2; p * p <= MAXN; p++) {
 
        // If prime[p] is not changed, then it is a prime
        if (prime[p] == true) {
 
            // Update all multiples of p as non-prime
            for (var i = p * p; i <= MAXN; i += p)
                prime[i] = false;
        }
    }
}
 
// Find the common prime divisors
function common_prime( a, b)
{
 
    // Get the GCD of the given numbers
    var gcd = __gcd(a, b);
 
    // Find the prime divisors of the gcd
    for (var i = 2; i <= (gcd); i++) {
 
        // If i is prime and a divisor of gcd
        if (prime[i] && gcd % i == 0) {
            document.write( i + " ");
       }
    }
}
 
 
SieveOfEratosthenes();
var a = 6, b = 12;
common_prime(a, b);
 
//This code is contributed by SoumikModnal
</script>
Output: 
2 3

 




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