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: 16Input: 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++ 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 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 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# 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 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. ?> |
<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> |
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.