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.
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++
#include <iostream>
using namespace std;
int gcd(int a, int b)
{
if (a == 0)
return b;
if (b == 0)
return a;
if (a == b)
return a;
if (a > b)
return gcd(a-b, b);
return gcd(a, b-a);
}
int main()
{
int a = 98, b = 56;
cout<<"GCD of "<<a<<" and "<<b<<" is "<<gcd(a, b);
return 0;
}
|
C
#include <stdio.h>
int gcd(int a, int b)
{
if (a == 0)
return b;
if (b == 0)
return a;
if (a == b)
return a;
if (a > b)
return gcd(a-b, b);
return gcd(a, b-a);
}
int main()
{
int a = 98, b = 56;
printf("GCD of %d and %d is %d ", a, b, gcd(a, b));
return 0;
}
|
Java
class Test
{
static int gcd(int a, int b)
{
if (a == 0)
return b;
if (b == 0)
return a;
if (a == b)
return a;
if (a > b)
return gcd(a-b, b);
return gcd(a, b-a);
}
public static void main(String[] args)
{
int a = 98, b = 56;
System.out.println("GCD of " + a +" and " + b + " is " + gcd(a, b));
}
}
|
Python3
def gcd(a,b):
if (a == 0):
return b
if (b == 0):
return a
if (a == b):
return a
if (a > b):
return gcd(a-b, b)
return gcd(a, b-a)
a = 98
b = 56
if(gcd(a, b)):
print('GCD of', a, 'and', b, 'is', gcd(a, b))
else:
print('not found')
|
C#
using System;
class GFG {
static int gcd(int a, int b)
{
if (a == 0)
return b;
if (b == 0)
return a;
if (a == b)
return a;
if (a > b)
return gcd(a - b, b);
return gcd(a, b - a);
}
public static void Main()
{
int a = 98, b = 56;
Console.WriteLine("GCD of "
+ a +" and " + b + " is "
+ gcd(a, b));
}
}
|
PHP
<?php
function gcd($a, $b)
{
if ($a == 0)
return $b;
if ($b == 0)
return $a;
if($a == $b)
return $a ;
if($a > $b)
return gcd( $a-$b , $b ) ;
return gcd( $a , $b-$a ) ;
}
$a = 98 ;
$b = 56 ;
echo "GCD of $a and $b is ", gcd($a , $b) ;
?>
|
Output:
GCD of 98 and 56 is 14
A more efficient solution is to use modulo operator in Euclidean algorithm .
C++
#include <iostream>
using namespace std;
int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
int main()
{
int a = 98, b = 56;
cout<<"GCD of "<<a<<" and "<<b<<" is "<<gcd(a, b);
return 0;
}
|
C
#include <stdio.h>
int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
int main()
{
int a = 98, b = 56;
printf("GCD of %d and %d is %d ", a, b, gcd(a, b));
return 0;
}
|
Java
class Test
{
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
public static void main(String[] args)
{
int a = 98, b = 56;
System.out.println("GCD of " + a +" and " + b + " is " + gcd(a, b));
}
}
|
Python3
def gcd(a,b):
if (b == 0):
return a
return gcd(b, a%b)
a = 98
b = 56
if(gcd(a, b)):
print('GCD of', a, 'and', b, 'is', gcd(a, b))
else:
print('not found')
|
C#
using System;
class GFG {
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
public static void Main()
{
int a = 98, b = 56;
Console.WriteLine("GCD of "
+ a +" and " + b + " is "
+ gcd(a, b));
}
}
|
PHP
<?php
function gcd($a, $b)
{
if($b==0)
return $a ;
return gcd( $b , $a % $b ) ;
}
$a = 98 ;
$b = 56 ;
echo "GCD of $a and $b is ", gcd($a , $b) ;
?>
|
Output:
GCD of 98 and 56 is 14
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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