LCM of given array elements
Given an array of n numbers, find LCM of it.
Input : {1, 2, 8, 3} Output : 24 Input : {2, 7, 3, 9, 4} Output : 252
The above relation only holds for two numbers,
The idea here is to extend our relation for more than 2 numbers. Let’s say we have an array arr[] that contains n elements whose LCM needed to be calculated.
The main steps of our algorithm are:
 Initialize ans = arr[0].
 Iterate over all the elements of the array i.e. from i = 1 to i = n1
At the ith iteration ans = LCM(arr[0], arr[1], …….., arr[i1]). This can be done easily as LCM(arr[0], arr[1], …., arr[i]) = LCM(ans, arr[i]). Thus at i’th iteration we just have to do ans = LCM(ans, arr[i]) = ans x arr[i] / gcd(ans, arr[i])
Below is a implementation in C++ of above algorithm:
 C++

// C++ program to find LCM of n elements
#include <bits/stdc++.h>
using
namespace
std;
typedef
long
long
int
ll;
// Utility function to find GCD of 'a' and 'b'
int
gcd(
int
a,
int
b)
{
if
(b==0)
return
a;
return
gcd(b, a%b);
}
// Returns LCM of array elements
ll findlcm(
int
arr[],
int
n)
{
// Initialize result
ll ans = arr[0];
// ans contains LCM of arr[0],..arr[i]
// after i'th iteration,
for
(
int
i=1; i<n; i++)
ans = ( ((arr[i]*ans)) /
(gcd(arr[i], ans)) );
return
ans;
}
// Driver Code
int
main()
{
int
arr[] = {2, 7, 3, 9, 4};
int
n =
sizeof
(arr)/
sizeof
(arr[0]);
printf
(
"%lld"
, findlcm(arr, n));
return
0;
}
Output :
252
Disclaimer: This does not belong to TechCodeBit, its an article taken from the below
source and credits.
source and credits: http://www.geeksforgeeks.org
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