Common Divisors of Two Numbers

Given two integer numbers, the task is to find count of all common divisors of given numbers?

Examples : 

Input : a = 12, b = 24
Output: 6
// all common divisors are 1, 2, 3, 
// 4, 6 and 12

Input : a = 3, b = 17
Output: 1
// all common divisors are 1

Input : a = 20, b = 36
Output: 3
// all common divisors are 1, 2, 4

It is recommended to refer all divisors of a given number as a prerequisite of this article. 

Naive Solution 
A simple solution is to first find all divisors of first number and store them in an array or hash. Then find common divisors of second number and store them. Finally print common elements of two stored arrays or hash. The key is that the magnitude of powers of prime factors of a divisor should be equal to the minimum power of two prime factors of a and b.

  • Find the prime factors of a using prime factorization.
  • Find the count of each prime factor of a and store it in a Hashmap.
  • Prime factorize b using distinct prime factors of a.
  • Then the total number of divisors would be equal to the product of (count + 1) 
    of each factor.
  • count is the minimum of counts of each prime factors of a and b.
  • This gives the count of all divisors of a and b.

C++

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// C++ implementation of program
#include <bits/stdc++.h>
using namespace std;
 
// Map to store the count of each
// prime factor of a
map<int, int> ma;
 
// Function that calculate the count of
// each prime factor of a number
void primeFactorize(int a)
{
    for(int i = 2; i * i <= a; i += 2)
    {
        int cnt = 0;
        while (a % i == 0)
        {
            cnt++;
            a /= i;
        }
        ma[i] = cnt;
    }
    if (a > 1)
    {
        ma[a] = 1;
    }
}
 
// Function to calculate all common
// divisors of two given numbers
// a, b --> input integer numbers
int commDiv(int a, int b)
{
     
    // Find count of each prime factor of a
    primeFactorize(a);
 
    // stores number of common divisors
    int res = 1;
 
    // Find the count of prime factors
    // of b using distinct prime factors of a
    for(auto m = ma.begin();
             m != ma.end(); m++)
    {
        int cnt = 0;
        int key = m->first;
        int value = m->second;
 
        while (b % key == 0)
        {
            b /= key;
            cnt++;
        }
 
        // Prime factor of common divisor
        // has minimum cnt of both a and b
        res *= (min(cnt, value) + 1);
    }
    return res;
}
 
// Driver code   
int main()
{
    int a = 12, b = 24;
     
    cout << commDiv(a, b) << endl;
     
    return 0;
}
 
// This code is contributed by divyeshrabadiya07

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Java

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// Java implementation of program
import java.util.*;
import java.io.*;
 
class GFG {
    // map to store the count of each prime factor of a
    static HashMap<Integer, Integer> ma = new HashMap<>();
 
    // method that calculate the count of
    // each prime factor of a number
    static void primeFactorize(int a)
    {
        for (int i = 2; i * i <= a; i += 2) {
            int cnt = 0;
            while (a % i == 0) {
                cnt++;
                a /= i;
            }
            ma.put(i, cnt);
        }
        if (a > 1)
            ma.put(a, 1);
    }
 
    // method to calculate all common divisors
    // of two given numbers
    // a, b --> input integer numbers
    static int commDiv(int a, int b)
    {
        // Find count of each prime factor of a
        primeFactorize(a);
 
        // stores number of common divisors
        int res = 1;
 
        // Find the count of prime factors of b using
        // distinct prime factors of a
        for (Map.Entry<Integer, Integer> m : ma.entrySet()) {
            int cnt = 0;
 
            int key = m.getKey();
            int value = m.getValue();
 
            while (b % key == 0) {
                b /= key;
                cnt++;
            }
 
            // prime factor of common divisor
            // has minimum cnt of both a and b
            res *= (Math.min(cnt, value) + 1);
        }
        return res;
    }
 
    // Driver method
    public static void main(String args[])
    {
        int a = 12, b = 24;
        System.out.println(commDiv(a, b));
    }
}

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C#

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// C# implementation of program
using System;
using System.Collections.Generic;
 
class GFG{
   
// Map to store the count of each
// prime factor of a
static Dictionary<int,
                  int> ma = new Dictionary<int,
                                           int>();
 
// Function that calculate the count of
// each prime factor of a number
static void primeFactorize(int a)
{
    for(int i = 2; i * i <= a; i += 2)
    {
        int cnt = 0;
        while (a % i == 0)
        {
            cnt++;
            a /= i;
        }
        ma.Add(i, cnt);
    }
     
    if (a > 1)
        ma.Add(a, 1);
}
 
// Function to calculate all common
// divisors of two given numbers
// a, b --> input integer numbers
static int commDiv(int a, int b)
{
     
    // Find count of each prime factor of a
    primeFactorize(a);
     
    // Stores number of common divisors
    int res = 1;
     
    // Find the count of prime factors
    // of b using distinct prime factors of a
    foreach(KeyValuePair<int, int> m in ma)
    {
        int cnt = 0;
        int key = m.Key;
        int value = m.Value;
         
        while (b % key == 0)
        {
            b /= key;
            cnt++;
        }
 
        // Prime factor of common divisor
        // has minimum cnt of both a and b
        res *= (Math.Min(cnt, value) + 1);
    }
    return res;
}
 
// Driver code   
static void Main()
{
    int a = 12, b = 24;
     
    Console.WriteLine(commDiv(a, b));
}
}
 
// This code is contributed by divyesh072019

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Output: 



6

Efficient Solution – 
A better solution is to calculate the greatest common divisor (gcd) of given two numbers, and then count divisors of that gcd. 

C++

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// C++ implementation of program
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate gcd of two numbers
int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}
 
// Function to calculate all common divisors
// of two given numbers
// a, b --> input integer numbers
int commDiv(int a, int b)
{
    // find gcd of a, b
    int n = gcd(a, b);
 
    // Count divisors of n.
    int result = 0;
    for (int i = 1; i <= sqrt(n); i++) {
        // if 'i' is factor of n
        if (n % i == 0) {
            // check if divisors are equal
            if (n / i == i)
                result += 1;
            else
                result += 2;
        }
    }
    return result;
}
 
// Driver program to run the case
int main()
{
    int a = 12, b = 24;
    cout << commDiv(a, b);
    return 0;
}

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Java

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// Java implementation of program
 
class Test {
    // method to calculate gcd of two numbers
    static int gcd(int a, int b)
    {
        if (a == 0)
            return b;
 
        return gcd(b % a, a);
    }
    // method to calculate all common divisors
    // of two given numbers
    // a, b --> input integer numbers
    static int commDiv(int a, int b)
    {
        // find gcd of a, b
        int n = gcd(a, b);
 
        // Count divisors of n.
        int result = 0;
        for (int i = 1; i <= Math.sqrt(n); i++) {
            // if 'i' is factor of n
            if (n % i == 0) {
                // check if divisors are equal
                if (n / i == i)
                    result += 1;
                else
                    result += 2;
            }
        }
        return result;
    }
 
    // Driver method
    public static void main(String args[])
    {
        int a = 12, b = 24;
        System.out.println(commDiv(a, b));
    }
}

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Python3

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# Python implementation of program
from math import sqrt
 
 
# Function to calculate gcd of two numbers
def gcd(a, b):
     
    if a == 0:
        return b
    return gcd(b % a, a)
   
# Function to calculate all common divisors
# of two given numbers
# a, b --> input integer numbers
def commDiv(a, b):
     
    # find GCD of a, b
    n = gcd(a, b)
 
    # Count divisors of n
    result = 0
    for i in range(1,int(sqrt(n))+1):
 
        # if i is a factor of n
        if n % i == 0:
 
            # check if divisors are equal
            if n/i == i:
                result += 1
            else:
                result += 2
                 
    return result
 
# Driver program to run the case
if __name__ == "__main__":
    a = 12
    b = 24;
    print(commDiv(a, b))

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C#

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// C# implementation of program
using System;
 
class GFG {
 
    // method to calculate gcd
    // of two numbers
    static int gcd(int a, int b)
    {
        if (a == 0)
            return b;
 
        return gcd(b % a, a);
    }
 
    // method to calculate all
    // common divisors of two
    // given numbers a, b -->
    // input integer numbers
    static int commDiv(int a, int b)
    {
 
        // find gcd of a, b
        int n = gcd(a, b);
 
        // Count divisors of n.
        int result = 0;
        for (int i = 1; i <= Math.Sqrt(n); i++) {
 
            // if 'i' is factor of n
            if (n % i == 0) {
 
                // check if divisors are equal
                if (n / i == i)
                    result += 1;
                else
                    result += 2;
            }
        }
 
        return result;
    }
 
    // Driver method
    public static void Main(String[] args)
    {
 
        int a = 12, b = 24;
 
        Console.Write(commDiv(a, b));
    }
}
 
// This code contributed by parashar.

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PHP

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<?php
// PHP implementation of program
 
// Function to calculate
// gcd of two numbers
function gcd($a, $b)
{
    if ($a == 0)
        return $b;
    return gcd($b % $a, $a);
}
 
// Function to calculate all common
// divisors of two given numbers
// a, b --> input integer numbers
function commDiv($a, $b)
{
    // find gcd of a, b
    $n = gcd($a, $b);
 
    // Count divisors of n.
    $result = 0;
    for ($i = 1; $i <= sqrt($n);
                 $i++)
    {
        // if 'i' is factor of n
        if ($n % $i == 0)
        {
            // check if divisors
            // are equal
            if ($n / $i == $i)
                $result += 1;
            else
                $result += 2;
        }
    }
    return $result;
}
 
// Driver Code
$a = 12; $b = 24;
echo(commDiv($a, $b));
 
// This code is contributed by Ajit.
?>

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Output :  

6

Time complexity: O(log(n)+ n1/2) where n is the gcd of two numbers.

This article is contributed by Shashank Mishra ( Gullu ). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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