Finding LCM using GCD is explained here but here the task is to find LCM without first calculating GCD.
Examples:
Input: 7, 5 Output: 35 Input: 2, 6 Output: 6
The approach is to start with the largest of the 2 numbers and keep incrementing the larger number by itself till smaller number perfectly divides the resultant.
C++
// C++ program to find LCM of 2 numbers // without using GCD #include <bits/stdc++.h> using namespace std;
// Function to return LCM of two numbers int findLCM( int a, int b)
{ int lar = max(a, b);
int small = min(a, b);
for ( int i = lar; ; i += lar) {
if (i % small == 0)
return i;
}
} // Driver program to test above function int main()
{ int a = 5, b = 7;
cout << "LCM of " << a << " and "
<< b << " is " << findLCM(a, b);
return 0;
} |
Java
// Java program to find LCM of 2 numbers // without using GCD import java.io.*;
import java.lang.*;
class GfG {
// Function to return LCM of two numbers
public static int findLCM( int a, int b)
{
int lar = Math.max(a, b);
int small = Math.min(a, b);
for ( int i = lar; ; i += lar) {
if (i % small == 0 )
return i;
}
}
// Driver program to test above function
public static void main(String [] argc)
{
int a = 5 , b = 7 ;
System.out.println( "LCM of " + a + " and "
+ b + " is " + findLCM(a, b));
}
} // This dose is contributed by Sagar Shukla. |
Python 3
# Python 3 program to find # LCM of 2 numbers without # using GCD import sys
# Function to return # LCM of two numbers def findLCM(a, b):
lar = max (a, b)
small = min (a, b)
i = lar
while ( 1 ) :
if (i % small = = 0 ):
return i
i + = lar
# Driver Code a = 5
b = 7
print ( "LCM of " , a , " and " ,
b , " is " ,
findLCM(a, b), sep = "")
# This code is contributed # by Smitha |
C#
// C# program to find // LCM of 2 numbers // without using GCD using System;
class GfG
{ // Function to return
// LCM of two numbers
public static int findLCM( int a,
int b)
{
int lar = Math.Max(a, b);
int small = Math.Min(a, b);
for ( int i = lar; ; i += lar)
{
if (i % small == 0)
return i;
}
}
// Driver Code
public static void Main()
{
int a = 5, b = 7;
Console.WriteLine( "LCM of " + a +
" and " + b +
" is " +
findLCM(a, b));
}
} // This code is contributed by anuj_67. |
PHP
<?php // PHP program to find // LCM of 2 numbers // without using GCD // Function to return // LCM of two numbers function findLCM( $a , $b )
{ $lar = max( $a , $b );
$small = min( $a , $b );
for ( $i = $lar ; ; $i += $lar )
{
if ( $i % $small == 0)
return $i ;
}
} // Driver Code $a = 5;
$b = 7;
echo "LCM of " , $a , " and " ,
$b , " is " ,
findLCM( $a , $b );
// This code is contributed // by Smitha ?> |
Javascript
<script> // javascript program to find LCM of 2 numbers // without using GCD // Function to return LCM of two numbers
function findLCM(a , b) {
var lar = Math.max(a, b);
var small = Math.min(a, b);
for (i = lar;; i += lar) {
if (i % small == 0)
return i;
}
}
// Driver program to test above function
var a = 5, b = 7;
document.write( "LCM of " + a + " and " + b + " is " + findLCM(a, b));
// This code contributed by umadevi9616 </script> |
Output:
LCM of 5 and 7 is 35
Time Complexity: O(max(a, b))
Auxiliary Space: O(1)