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Program to find HCF (Highest Common Factor) of 2 Numbers

HCF (Highest Common Factor) or GCD (Greatest Common Divisor) 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.

A simple solution is to find all prime factors of both numbers, then find the intersection of all factors present in both numbers. Finally, return the 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 && b == 0)``        ``return` `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 = 0, b = 56;``    ``cout << ``"GCD of "` `<< a << ``" and "` `<< b <<  ``" is "` `<< gcd(a, b);``    ``return` `0;``}` `// This code is contributed by shivanisinghss2110`

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 && b == 0)``        ``return` `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 = 0, 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``import` `java.io.*;` `class` `Test {``    ``// Recursive function to return gcd of a and b``    ``static` `int` `gcd(``int` `a, ``int` `b)``    ``{``        ``// Everything divides 0``        ``if` `(a == ``0` `&& b == ``0``)``            ``return` `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` `and` `b ``=``=` `0``):``        ``return` `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 && b == 0)``            ``return` `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

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Output

`GCD of 0 and 56 is 56`

Time Complexity: O(max(a,b)), where a and b are the given two integers.
Auxiliary Space: O(max(a,b)), where a and b are the given two integers.

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``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;``    ``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``import` `java.io.*;` `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

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Javascript

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Output

`GCD of 98 and 56 is 14`

Time Complexity: O(log(max(a,b)), where a and b are the given two integers.
Auxiliary Space: O(log(max(a,b)), where a and b are the given two integers.

Please refer GCD of more than two (or array) numbers to find HCF of more than two numbers.