Program to find GCD or HCF of two numbers - GeeksforGeeks

03 Feb, 2022
GCD (Greatest Common Divisor) or HCF (Highest Common Factor) of two numbers is the largest number that divides both of them. 
For example GCD of 20 and 28 is 4 and GCD of 98 and 56 is 14. For solution  suppose a=98 & b=56  
a>b so put a= a-b and b is  remain same  so  a=98-56=42  & b= 56 . Now b>a  so  b=b-a and  a is same 
b= 56-42 = 14 & a= 42   . 42 is  3 times of 14  so HCF is 14  . likewise  a=36  & b=60  ,here b>a                        so b = 24 & a= 36  now a>b so a= 12 & b= 24  . 12 is HCF of 36 and 60 .  This  concept  is  always  satisfying. 
 A simple solution is to find all prime factors of both numbers, then find intersection of all factors present in both numbers. Finally return product of elements in the intersection.An efficient solution is to use Euclidean algorithm which is the main algorithm used for this purpose. The idea is, GCD of two numbers doesn’t change if smaller number is subtracted from a bigger number. 
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// C++ program to find GCD of two numbers#include <iostream>using namespace std;// Recursive function to return gcd of a and bint gcd(int a, int b){    // Everything divides 0    if (a == 0)       return b;    if (b == 0)       return a;      // base case    if (a == b)        return a;      // a is greater    if (a > b)        return gcd(a-b, b);    return gcd(a, b-a);}  // Driver program to test above functionint main(){    int a = 98, b = 56;    cout<<"GCD of "<<a<<" and "<<b<<" is "<<gcd(a, b);    return 0;}
// C program to find GCD of two numbers#include <stdio.h> // Recursive function to return gcd of a and bint gcd(int a, int b){    // Everything divides 0    if (a == 0)       return b;    if (b == 0)       return a;     // base case    if (a == b)        return a;     // a is greater    if (a > b)        return gcd(a-b, b);    return gcd(a, b-a);} // Driver program to test above functionint main(){    int a = 98, b = 56;    printf("GCD of %d and %d is %d ", a, b, gcd(a, b));    return 0;}
// Java program to find GCD of two numbersclass Test{    // Recursive function to return gcd of a and b    static int gcd(int a, int b)    {        // Everything divides 0        if (a == 0)          return b;        if (b == 0)          return a;              // base case        if (a == b)            return a;              // a is greater        if (a > b)            return gcd(a-b, b);        return gcd(a, b-a);    }         // Driver method    public static void main(String[] args)    {        int a = 98, b = 56;        System.out.println("GCD of " + a +" and " + b + " is " + gcd(a, b));    }}
# Recursive function to return gcd of a and bdef gcd(a,b):         # Everything divides 0    if (a == 0):        return b    if (b == 0):        return a     # base case    if (a == b):        return a     # a is greater    if (a > b):        return gcd(a-b, b)    return gcd(a, b-a) # Driver program to test above functiona = 98b = 56if(gcd(a, b)):    print('GCD of', a, 'and', b, 'is', gcd(a, b))else:    print('not found') # This code is contributed by Danish Raza
// C# program to find GCD of two// numbersusing System; class GFG {         // Recursive function to return    // gcd of a and b    static int gcd(int a, int b)    {                 // Everything divides 0        if (a == 0)          return b;        if (b == 0)          return a;             // base case        if (a == b)            return a;             // a is greater        if (a > b)            return gcd(a - b, b);                     return gcd(a, b - a);    }         // Driver method    public static void Main()    {        int a = 98, b = 56;        Console.WriteLine("GCD of "          + a +" and " + b + " is "                      + gcd(a, b));    }} // This code is contributed by anuj_67.
<?php// PHP program to find GCD// of two numbers // Recursive function to// return gcd of a and bfunction gcd($a, $b){     // Everything divides 0    if ($a == 0)       return $b;    if ($b == 0)       return $a;     // base case    if($a == $b)        return $a ;         // a is greater    if($a > $b)        return gcd( $a-$b , $b ) ;     return gcd( $a , $b-$a ) ;} // Driver code$a = 98 ;$b = 56 ; echo "GCD of $a and $b is ", gcd($a , $b) ; // This code is contributed by Anivesh Tiwari?>
<script> // Javascript program to find GCD of two numbers // Recursive function to return gcd of a and bfunction gcd(a, b){    // Everything divides 0    if (a == 0)    return b;    if (b == 0)    return a;     // base case    if (a == b)        return a;     // a is greater    if (a > b)        return gcd(a-b, b);    return gcd(a, b-a);} // Driver program to test above function     let a = 98, b = 56;    document.write("GCD of "+ a + " and " + b + " is " + gcd(a, b));     // This code is contributed by Mayank Tyagi </script>
Output: 
GCD of 98 and 56 is 14
Time Complexity: O(max(a,b))
Auxiliary Space: O(max(a,b))
Dynamic Programming Approach (Top Down Using Memoization) :
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// C++ program to find GCD of two numbers#include <bits/stdc++.h>using namespace std; int static dp[1001][1001]; // Function to return gcd of a and bint gcd(int a, int b){    // Everything divides 0    if (a == 0)        return b;    if (b == 0)        return a;     // base case    if (a == b)        return a;         // if a value is already    // present in dp    if(dp[a][b] != -1)        return dp[a][b];     // a is greater    if (a > b)        dp[a][b] = gcd(a-b, b);         // b is greater    else        dp[a][b] = gcd(a, b-a);         // return dp    return dp[a][b];} // Driver program to test above functionint main(){    int a = 98, b = 56;    memset(dp, -1, sizeof(dp));    cout<<"GCD of "<<a<<" and "<<b<<" is "<<gcd(a, b);    return 0;}
// Java program to find GCD of two numbersimport java.util.*;public class GFG{    static int [][]dp = new int[1001][1001];       // Recursive function to return gcd of a and b    static int gcd(int a, int b)    {               // Everything divides 0        if (a == 0)          return b;        if (b == 0)          return a;              // base case        if (a == b)            return a;              // if a value is already    // present in dp    if(dp[a][b] != -1)        return dp[a][b];     // a is greater    if (a > b)        dp[a][b] = gcd(a-b, b);         // b is greater    else        dp[a][b] = gcd(a, b-a);         // return dp    return dp[a][b];    }         // Driver method    public static void main(String[] args)    {        for(int i = 0; i < 1001; i++) {            for(int j = 0; j < 1001; j++) {                dp[i][j] = -1;            }        }        int a = 98, b = 56;        System.out.println("GCD of " + a +" and " + b + " is " + gcd(a, b));    }} // This code is contributed by Samim Hossain Mondal.
# function to return gcd of a and b # Taking the matrix as globallydp = [[-1 for i in range(1001)] for j in range(1001)] def gcd(a,b):         # Everything divides 0    if (a == 0):        return b    if (b == 0):        return a     # base case    if (a == b):        return a         if(dp[a][b] != -1):        return dp[a][b]             # a is greater    if (a > b):        dp[a][b] = gcd(a-b, b)    else:        dp[a][b] = gcd(a, b-a)             return dp[a][b] # Driver program to test above functiona = 98b = 56if(gcd(a, b)):    print('GCD of', a, 'and', b, 'is', gcd(a, b))else:    print('not found') # This code is contributed by Samim Hossain Mondal.
// C# program to find GCD of two numbersusing System;class GFG{    static int [,]dp = new int[1001, 1001];       // Recursive function to return gcd of a and b    static int gcd(int a, int b)    {               // Everything divides 0        if (a == 0)          return b;        if (b == 0)          return a;              // base case        if (a == b)            return a;          // if a value is already    // present in dp    if(dp[a, b] != -1)        return dp[a, b];     // a is greater    if (a > b)        dp[a, b] = gcd(a-b, b);         // b is greater    else        dp[a, b] = gcd(a, b-a);         // return dp    return dp[a, b];    }         // Driver method    public static void Main()    {        for(int i = 0; i < 1001; i++) {            for(int j = 0; j < 1001; j++) {                dp[i, j] = -1;            }        }        int a = 98, b = 56;        Console.Write("GCD of " + a +" and " + b + " is " + gcd(a, b));    }} // This code is contributed by Samim Hossain Mondal.
//<script>// Javascript program to find GCD of two numbersvar dp = new Array(1001); // Loop to create 2D array using 1D arrayfor (var i = 0; i < dp.length; i++) {    dp[i] = new Array(1001);} // Function to return gcd of a and bfunction gcd(a, b){     // Everything divides 0    if (a == 0)        return b;    if (b == 0)        return a;     // base case    if (a == b)        return a;         // if a value is already    // present in dp    if(dp[a][b] != -1)        return dp[a][b];     // a is greater    if (a > b)        dp[a][b] = gcd(a-b, b);         // b is greater    else        dp[a][b] = gcd(a, b-a);         // return dp    return dp[a][b];} // Driver program to test above function    let a = 98, b = 56;         for(let i = 0; i < 1001; i++) {        for(let j = 0; j < 1001; j++) {            dp[i][j] = -1;        }    }    document.write("GCD of "+ a + " and " + b + " is " + gcd(a, b));     // This code is contributed by Samim Hossain Mondal </script>
GCD of 98 and 56 is 14
Time Complexity: O(max(a,b))
Auxiliary Space: O(1)
A more efficient solution is to use modulo operator in Euclidean algorithm.
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Java
Python3
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PHP
Javascript
// C++ program to find GCD of two numbers#include <iostream>using namespace std;// Recursive function to return gcd of a and b in single lineint gcd(int a, int b){    return b == 0 ? a : gcd(b, a % b);   }  // Driver program to test above functionint main(){    int a = 98, b = 56;    cout<<"GCD of "<<a<<" and "<<b<<" is "<<gcd(a, b);    return 0;}
// C program to find GCD of two numbers#include <stdio.h> // Recursive function to return gcd of a and bint gcd(int a, int b){    if (b == 0)        return a;    return gcd(b, a % b);} // Driver program to test above functionint main(){    int a = 98, b = 56;    printf("GCD of %d and %d is %d ", a, b, gcd(a, b));    return 0;}
// Java program to find GCD of two numbersclass Test{    // Recursive function to return gcd of a and b    static int gcd(int a, int b)    {      if (b == 0)        return a;      return gcd(b, a % b);    }         // Driver method    public static void main(String[] args)    {        int a = 98, b = 56;        System.out.println("GCD of " + a +" and " + b + " is " + gcd(a, b));    }}
# Recursive function to return gcd of a and bdef gcd(a,b):         # Everything divides 0    if (b == 0):         return a    return gcd(b, a%b) # Driver program to test above functiona = 98b = 56if(gcd(a, b)):    print('GCD of', a, 'and', b, 'is', gcd(a, b))else:    print('not found') # This code is contributed by Danish Raza
// C# program to find GCD of two// numbersusing System; class GFG {         // Recursive function to return    // gcd of a and b    static int gcd(int a, int b)    {            if (b == 0)          return a;       return gcd(b, a % b);    }         // Driver method    public static void Main()    {        int a = 98, b = 56;        Console.WriteLine("GCD of "          + a +" and " + b + " is "                      + gcd(a, b));    }} // This code is contributed by anuj_67.
<?php// PHP program to find GCD// of two numbers // Recursive function to// return gcd of a and bfunction gcd($a, $b){    // Everything divides 0    if($b==0)        return $a ;     return gcd( $b , $a % $b ) ;} // Driver code$a = 98 ;$b = 56 ; echo "GCD of $a and $b is ", gcd($a , $b) ; // This code is contributed by Anivesh Tiwari?>
<script> // Javascript program to find GCD of two number // Recursive function to return gcd of a and b function gcd(a, b){     // Everything divides 0  if(b == 0){    return a;  }     return gcd(b, a % b);} // Driver codelet a = 98;let b = 56; document.write(`GCD of ${a} and ${b} is ${gcd(a, b)}`); // This code is contributed by _saurabh_jaiswal </script>
Output: 
GCD of 98 and 56 is 14
Time Complexity: O(log(max(a,b))
Auxiliary Space: O(log(max(a,b))
The time complexity for the above algorithm is O(log(max(a,b))) the derivation for this is obtained from the analysis of the worst-case scenario. What we do is we ask what are the 2 least numbers that take 1 step, those would be (1,1). If we want to increase the number of steps to 2 while keeping the numbers as low as possible as we can take the numbers to be (1,2). Similarly, for 3 steps, the numbers would be (2,3), 4 would be (3,5), 5 would be (5,8). So we can notice a pattern here, for the nth step the numbers would be (fib(n),fib(n+1)).  So the worst-case time complexity would be O(n) where a>= fib(n) and b>= fib(n+1). 
Now Fibonacci series is an exponentially growing series where the ratio of nth/(n-1)th term approaches (sqrt(5)-1)/2 which is also called the golden ratio. So we can see that the time complexity of the algorithm increases linearly as the terms grow exponentially hence the time complexity would be log(max(a,b)).
Please refer GCD of more than two (or array) numbers to find HCF of more than two numbers.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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