# Program to find GCD or HCF of two numbers

• Difficulty Level : Easy
• Last Updated : 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.

## C++

 `// C++ program to find GCD of two numbers``#include ``using` `namespace` `std;``// Recursive function to return gcd of a and b``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 program to test above function``int` `main()``{``    ``int` `a = 98, b = 56;``    ``cout<<``"GCD of "``<

## C

 `// C program to find GCD of two numbers``#include ` `// Recursive function to return gcd of a and b``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 program to test above function``int` `main()``{``    ``int` `a = 98, b = 56;``    ``printf``(``"GCD of %d and %d is %d "``, a, b, gcd(a, b));``    ``return` `0;``}`

## Java

 `// Java program to find GCD of two numbers``class` `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));``    ``}``}`

## Python3

 `# Recursive function to return gcd of a and b``def` `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``a ``=` `98``b ``=` `56``if``(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#

 `// C# program to find GCD of two``// numbers``using` `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

 ` ``\$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``?>`

## Javascript

 ``

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) :

## C++

 `// C++ program to find GCD of two numbers``#include ``using` `namespace` `std;` `int` `static` `dp[1001][1001];` `// Function to return gcd of a and b``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 program to test above function``int` `main()``{``    ``int` `a = 98, b = 56;``    ``memset``(dp, -1, ``sizeof``(dp));``    ``cout<<``"GCD of "``<

## Java

 `// Java program to find GCD of two numbers``import` `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.`

## Python3

 `# function to return gcd of a and b` `# Taking the matrix as globally``dp ``=` `[[``-``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 function``a ``=` `98``b ``=` `56``if``(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#

 `// C# program to find GCD of two numbers``using` `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.`

## Javascript

 `//`
Output
`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.

## C++

 `// C++ program to find GCD of two numbers``#include ``using` `namespace` `std;``// Recursive function to return gcd of a and b in single line``int` `gcd(``int` `a, ``int` `b)``{``    ``return` `b == 0 ? a : gcd(b, a % b);   ``}`` ` `// Driver program to test above function``int` `main()``{``    ``int` `a = 98, b = 56;``    ``cout<<``"GCD of "``<

## C

 `// C program to find GCD of two numbers``#include ` `// Recursive function to return gcd of a and b``int` `gcd(``int` `a, ``int` `b)``{``    ``if` `(b == 0)``        ``return` `a;``    ``return` `gcd(b, a % b);``}` `// Driver program to test above function``int` `main()``{``    ``int` `a = 98, b = 56;``    ``printf``(``"GCD of %d and %d is %d "``, a, b, gcd(a, b));``    ``return` `0;``}`

## Java

 `// Java program to find GCD of two numbers``class` `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));``    ``}``}`

## Python3

 `# Recursive function to return gcd of a and b``def` `gcd(a,b):``    ` `    ``# Everything divides 0``    ``if` `(b ``=``=` `0``):``         ``return` `a``    ``return` `gcd(b, a``%``b)` `# Driver program to test above function``a ``=` `98``b ``=` `56``if``(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#

 `// C# program to find GCD of two``// numbers``using` `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

 ``

## Javascript

 ``

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.
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