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```

We know,

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:

1. Initialize ans = arr[0].
2. Iterate over all the elements of the array i.e. from i = 1 to i = n-1
At the ith iteration ans = LCM(arr[0], arr[1], …….., arr[i-1]). 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++ 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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