Find HCF of two numbers without using recursion or Euclidean algorithm

Given two integer x and y, the task is to find the HCF of the numbers without using recursion or Euclidean method.

Examples:

Input: x = 16, y = 32
Output: 16

Input: x = 12, y = 15
Output: 3



Approach: HCF of two numbers is the greatest number which can divide both the numbers. If the smaller of the two numbers can divide the larger number then the HCF is the smaller number. Else starting from (smaller / 2) to 1 check whether the current element divides both the numbers . If yes, then it is the required HCF.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <iostream>
using namespace std;
  
// Function to return the HCF of x and y
int getHCF(int x, int y)
{
  
    // Minimum of the two numbers
    int minimum = min(x, y);
  
    // If both the numbers are divisible
    // by the minimum of these two then
    // the HCF is equal to the minimum
    if (x % minimum == 0 && y % minimum == 0)
        return minimum;
  
    // Highest number between 2 and minimum/2
    // which can divide both the numbers
    // is the required HCF
    for (int i = minimum / 2; i >= 2; i--) {
  
        // If both the numbers
        // are divisible by i
        if (x % i == 0 && y % i == 0)
            return i;
    }
  
    // 1 divides every number
    return 1;
}
  
// Driver code
int main()
{
    int x = 16, y = 32;
    cout << getHCF(x, y);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the approach
class GFG
{
      
// Function to return the HCF of x and y
static int getHCF(int x, int y)
{
  
    // Minimum of the two numbers
    int minimum = Math.min(x, y);
  
    // If both the numbers are divisible
    // by the minimum of these two then
    // the HCF is equal to the minimum
    if (x % minimum == 0 && y % minimum == 0)
        return minimum;
  
    // Highest number between 2 and minimum/2
    // which can divide both the numbers
    // is the required HCF
    for (int i = minimum / 2; i >= 2; i--)
    {
  
        // If both the numbers
        // are divisible by i
        if (x % i == 0 && y % i == 0)
            return i;
    }
  
    // 1 divides every number
    return 1;
}
  
// Driver code
public static void main(String[] args)
{
    int x = 16, y = 32;
    System.out.println(getHCF(x, y));
}
}
  
// This code is contributed by Code_Mech.

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the approach
  
# Function to return the HCF of x and y
def getHCF(x, y):
  
    # Minimum of the two numbers
    minimum = min(x, y)
  
    # If both the numbers are divisible
    # by the minimum of these two then
    # the HCF is equal to the minimum
    if (x % minimum == 0 and y % minimum == 0):
        return minimum
  
    # Highest number between 2 and minimum/2
    # which can divide both the numbers
    # is the required HCF
    for i in range(minimum // 2, 1, -1):
          
        # If both the numbers are divisible by i
        if (x % i == 0 and y % i == 0):
            return i
  
    # 1 divides every number
    return 1
  
# Driver code
x, y = 16, 32
print(getHCF(x, y))
  
# This code is contributed by mohit kumar

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the approach
using System;
  
class GFG
{
      
// Function to return the HCF of x and y
static int getHCF(int x, int y)
{
  
    // Minimum of the two numbers
    int minimum = Math.Min(x, y);
  
    // If both the numbers are divisible
    // by the minimum of these two then
    // the HCF is equal to the minimum
    if (x % minimum == 0 && y % minimum == 0)
        return minimum;
  
    // Highest number between 2 and minimum/2
    // which can divide both the numbers
    // is the required HCF
    for (int i = minimum / 2; i >= 2; i--)
    {
  
        // If both the numbers
        // are divisible by i
        if (x % i == 0 && y % i == 0)
            return i;
    }
  
    // 1 divides every number
    return 1;
}
  
// Driver code
static void Main()
{
    int x = 16, y = 32;
    Console.WriteLine(getHCF(x, y));
}
}
  
// This code is contributed by mits

chevron_right


Output:

16


My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.





Article Tags :
Practice Tags :


Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.