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# Java Program to Compute GCD

GCD (Greatest Common Divisor) of two given numbers A and B is the highest number that can divide both A and B completely, i.e., leaving remainder 0 in each case. GCD is also called HCF(Highest Common Factor). There are various approaches to find the GCD of two given numbers.

Approaches:

The GCD of the given two numbers A and B can be calculated using different approaches.

1. General method
2. Euclidean algorithm (by repeated subtraction)
3. Euclidean algorithm (by repeated division)

Examples:

```Input: 20, 30
Output: GCD(20, 30) = 10
Explanation: 10 is the highest integer which divides both 20 and 30 leaving 0 remainder

Input: 36, 37
Output: GCD(36, 37) = 1
Explanation: 36 and 37 don't have any factors in common except 1. So, 1 is the gcd of 36 and 37```

Note: gcd(A, B) = 1 if A, B are co-primes.

General Approach:

In the general approach of computing GCD, we actually implement the definition of GCD.

• First, find out all the factors of A and B individually.
• Then list out those factors which are common for both A and B.
• The highest of those common factors is the GCD of A and B.

Example:

```A = 20, B = 30
Factors of A : (1, 2, 4, 5, 10, 20)
Factors of B : (1, 2, 3, 5, 6, 10, 15, 30)
Common factors of A and B : (1, 2, 5, 10)
Highest of the Common factors (GCD) = 10```

It is clear that the GCD of 20 and 30 can’t be greater than 20. So we have to check for the numbers within the range 1 and 20. Also, we need the greatest of the divisors. So, iterate from backward to reduce computation time.

## Java

 `// Java program to compute GCD of``// two numbers using general``// approach``import` `java.io.*;` `class` `GFG {` `    ``// gcd() method, returns the GCD of a and b``    ``static` `int` `gcd(``int` `a, ``int` `b)``    ``{``        ``// stores minimum(a, b)``        ``int` `i;``        ``if` `(a < b)``            ``i = a;``        ``else``            ``i = b;` `        ``// take a loop iterating through smaller number to 1``        ``for` `(i = i; i > ``1``; i--) {` `            ``// check if the current value of i divides both``            ``// numbers with remainder 0 if yes, then i is``            ``// the GCD of a and b``            ``if` `(a % i == ``0` `&& b % i == ``0``)``                ``return` `i;``        ``}` `        ``// if there are no common factors for a and b other``        ``// than 1, then GCD of a and b is 1``        ``return` `1``;``    ``}``    ``// Driver method``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `a = ``30``, b = ``20``;` `        ``// calling gcd() method over``        ``// the integers 30 and 20``        ``System.out.println(``"GCD = "` `+ gcd(b, a));``    ``}``}`

Output

`GCD = 10`

Euclidean algorithm (repeated subtraction):

This approach is based on the principle that the GCD of two numbers A and B will be the same even if we replace the larger number with the difference between A and B. In this approach, we perform GCD operation on A and B repeatedly by replacing A with B and B with the difference(A, B) as long as the difference is greater than 0.

Example

```A = 30, B = 20
gcd(30, 20) -> gcd(A, B)
gcd(20, 30 - 20) = gcd(20,10) -> gcd(B,B-A)
gcd(30 - 20, 20 - (30 - 20)) = gcd(10, 10) -> gcd(B - A, B - (B - A))
gcd(10, 10 - 10) = gcd(10, 0)
here, the difference is 0
So stop the procedure. And 10 is the GCD of 30 and 20```

## Java

 `// Java program to compute GCD``// of two numbers using Euclid's``// repeated subtraction approach``import` `java.io.*;` `class` `GFG {` `    ``// gcd method returns the GCD of a and b``    ``static` `int` `gcd(``int` `a, ``int` `b)``    ``{``        ``// if b=0, a is the GCD``        ``if` `(b == ``0``)``            ``return` `a;` `        ``// call the gcd() method recursively by``        ``// replacing a with b and b with``        ``// difference(a,b) as long as b != 0``        ``else``            ``return` `gcd(b, Math.abs(a - b));``    ``}` `    ``// Driver method``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `a = ``30``, b = ``20``;` `        ``// calling gcd() over``        ``// integers 30 and 20``        ``System.out.println(``"GCD = "` `+ gcd(a, b));``    ``}``}`

Output

`GCD = 10`

Euclidean algorithm (repeated division):

This approach is similar to the repeated subtraction approach. But, in this approach, we replace B with the modulus of A and B instead of the difference.

Example :

```A = 30, B = 20
gcd(30, 20) -> gcd(A, B)
gcd(20, 30 % 20) = gcd(20, 10) -> gcd(B, A % B)
gcd(10, 20 % 10) = gcd(10, 10) -> gcd(A % B, B % (A % B))
gcd(10, 10 % 10) = gcd(10, 0)
here, the modulus became 0
So, stop the procedure. And 10 is the GCD of 30 and 20```

## Java

 `// Java program to compute GCD``// of two numbers using Euclid's``// repeated division approach``import` `java.io.*;``import` `java.util.*;` `class` `GFG {` `    ``// gcd method returns the GCD of a and b``    ``static` `int` `gcd(``int` `a, ``int` `b)``    ``{``        ``// if b=0, a is the GCD``        ``if` `(b == ``0``)``            ``return` `a;` `        ``// call the gcd() method recursively by``        ``// replacing a with b and b with``        ``// modulus(a,b) as long as b != 0``        ``else``            ``return` `gcd(b, a % b);``    ``}``    ``// Driver method``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `a = ``20``, b = ``30``;` `        ``// calling gcd() over``        ``// integers 30 and 20``        ``System.out.println(``"GCD = "` `+ gcd(a, b));``    ``}``}`

Output

`GCD = 10`

Euclid’s repeated division approach is most commonly used among all the approaches.

Time complexity: O(log(min(a,b)))

Auxiliary space: O(log(min(a,b)) for recursive call stack

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