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# Find HCF of two numbers without using recursion or Euclidean algorithm

• Difficulty Level : Medium
• Last Updated : 23 Jun, 2022

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 ``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

 `= 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

 ``

Output:

`16`

Time Complexity: O(min(x, y)), here x and y are two input parameters.
Auxiliary Space: O(1), since no extra space has been taken.

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