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,lcm

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 :


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