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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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. ?>

Output:

16

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