Common Divisors of Two Numbers - GeeksforGeeks

29 Jun, 2021
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++
Java
Python3
C#
Javascript
// C++ implementation of program#include <bits/stdc++.h>using namespace std; // Map to store the count of each// prime factor of amap<int, int> ma; // Function that calculate the count of// each prime factor of a numbervoid 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 numbersint 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
// Java implementation of programimport 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));    }}
# Python3 implementation of programimport math # Map to store the count of each# prime factor of ama = {} # Function that calculate the count of# each prime factor of a numberdef primeFactorize(a):         sqt = int(math.sqrt(a))    for i in range(2, sqt, 2):        cnt = 0                 while (a % i == 0):            cnt += 1            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 numbersdef commDiv(a, b):         # Find count of each prime factor of a    primeFactorize(a)         # stores number of common divisors    res = 1         # Find the count of prime factors    # of b using distinct prime factors of a    for key, value in ma.items():        cnt = 0                 while (b % key == 0):            b /= key            cnt += 1                     # Prime factor of common divisor        # has minimum cnt of both a and b        res *= (min(cnt, value) + 1)             return res     # Driver code   a = 12b = 24 print(commDiv(a, b)) # This code is contributed by Stream_Cipher
// C# implementation of programusing System;using System.Collections.Generic; class GFG{   // Map to store the count of each// prime factor of astatic Dictionary<int,                  int> ma = new Dictionary<int,                                           int>(); // Function that calculate the count of// each prime factor of a numberstatic 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 numbersstatic 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
<script>        // JavaScript implementation of program     // Map to store the count of each    // prime factor of a      let ma = new Map();     // Function that calculate the count of    // each prime factor of a number    function primeFactorize(a)    {        for(let i = 2; i * i <= a; i += 2)        {            let cnt = 0;            while (a % i == 0)            {                cnt++;                a = parseInt(a / i, 10);            }            ma.set(i, cnt);        }        if (a > 1)        {            ma.set(a, 1);        }    }     // Function to calculate all common    // divisors of two given numbers    // a, b --> input integer numbers    function commDiv(a,b)    {         // Find count of each prime factor of a        primeFactorize(a);         // stores number of common divisors        let res = 1;         // Find the count of prime factors        // of b using distinct prime factors of a        ma.forEach((values,keys)=>{            let cnt = 0;            let key = keys;            let value = values;             while (b % key == 0)            {                b = parseInt(b / key, 10);                cnt++;            }             // Prime factor of common divisor            // has minimum cnt of both a and b            res *= (Math.min(cnt, value) + 1);        })        return res;    }     // Driver code     let a = 12, b = 24;         document.write(commDiv(a, b));     </script>
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++
Java
Python3
C#
PHP
Javascript
// C++ implementation of program#include <bits/stdc++.h>using namespace std; // Function to calculate gcd of two numbersint 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 numbersint 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 caseint main(){    int a = 12, b = 24;    cout << commDiv(a, b);    return 0;}
// 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));    }}
# Python implementation of programfrom math import sqrt  # Function to calculate gcd of two numbersdef 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 numbersdef 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 caseif __name__ == "__main__":    a = 12    b = 24;    print(commDiv(a, b))
// C# implementation of programusing 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.
<?php// PHP implementation of program // Function to calculate// gcd of two numbersfunction gcd($a, $b){    if ($a == 0)        return $b;    return gcd($b % $a, $a);} // Function to calculate all common// divisors of two given numbers// a, b --> input integer numbersfunction commDiv($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.?>
<script>    // Javascript 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        let n = gcd(a, b);         // Count divisors of n.        let result = 0;        for (let 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;    }     let a = 12, b = 24;    document.write(commDiv(a, b));     </script>
Output :  
6
Time complexity: O(log(n)+ n1/2) where n is the gcd of two numbers.
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