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Find HCF of two numbers without using recursion or Euclidean algorithm
  • Difficulty Level : Easy
  • Last Updated : 07 Apr, 2021

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:
 

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++




// 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;
}

Java




// 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.

Python3




# 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

C#




// 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

PHP




<?php
// PHP implementation of the approach
 
// Function to return the HCF of x and y
function 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 &&
        $y % $minimum == 0)
        return $minimum;
 
    // Highest number between 2 and minimum/2
    // which can divide both the numbers
    // is the required HCF
    for ($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
$x = 16; $y = 32;
echo(getHCF($x, $y));
 
// This code is contributed
// by Code_Mech.
?>

Javascript




<script>
 
// Javascript implementation of the approach
 
// Function to return the HCF of x and y
function getHCF(x, y)
{
     
    // Minimum of the two numbers
    var 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(var 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
var x = 16, y = 32;
 
document.write(getHCF(x, y));
 
// This code is contributed by noob2000
 
</script>
Output: 
16

 

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