LCM of given array elements

Difficulty Level : Easy
Last Updated : 20 Oct, 2021

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

Recommended Practice

Number Game

Try It!

We know,
$LCM(a, b)=\frac{a*b}{gcd(a, b)}$
The above relation only holds for two numbers,
$LCM(a, b, c)\neq \frac{a*b*c}{gcd(a, b, c)}$
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 = 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 the implementation 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;
}

Java

// Java Program to find LCM of n elements
public class GFG {
     
    public static long lcm_of_array_elements(int[] element_array)
    {
        long lcm_of_array_elements = 1;
        int divisor = 2;
         
        while (true) {
            int counter = 0;
            boolean divisible = false;
             
            for (int i = 0; i < element_array.length; i++) {
 
                // lcm_of_array_elements (n1, n2, ... 0) = 0.
                // For negative number we convert into
                // positive and calculate lcm_of_array_elements.
 
                if (element_array[i] == 0) {
                    return 0;
                }
                else if (element_array[i] < 0) {
                    element_array[i] = element_array[i] * (-1);
                }
                if (element_array[i] == 1) {
                    counter++;
                }
 
                // Divide element_array by devisor if complete
                // division i.e. without remainder then replace
                // number with quotient; used for find next factor
                if (element_array[i] % divisor == 0) {
                    divisible = true;
                    element_array[i] = element_array[i] / divisor;
                }
            }
 
            // If divisor able to completely divide any number
            // from array multiply with lcm_of_array_elements
            // and store into lcm_of_array_elements and continue
            // to same divisor for next factor finding.
            // else increment divisor
            if (divisible) {
                lcm_of_array_elements = lcm_of_array_elements * divisor;
            }
            else {
                divisor++;
            }
 
            // Check if all element_array is 1 indicate 
            // we found all factors and terminate while loop.
            if (counter == element_array.length) {
                return lcm_of_array_elements;
            }
        }
    }
     
    // Driver Code
    public static void main(String[] args)
    {
        int[] element_array = { 2, 7, 3, 9, 4 };
        System.out.println(lcm_of_array_elements(element_array));
    }
}
 
// Code contributed by Mohit Gupta_OMG

Python

# Python Program to find LCM of n elements
 
def find_lcm(num1, num2):
    if(num1>num2):
        num = num1
        den = num2
    else:
        num = num2
        den = num1
    rem = num % den
    while(rem != 0):
        num = den
        den = rem
        rem = num % den
    gcd = den
    lcm = int(int(num1 * num2)/int(gcd))
    return lcm
     
l = [2, 7, 3, 9, 4]
 
num1 = l[0]
num2 = l[1]
lcm = find_lcm(num1, num2)
 
for i in range(2, len(l)):
    lcm = find_lcm(lcm, l[i])
     
print(lcm)
 
# Code contributed by Mohit Gupta_OMG

C#

// C# Program to find LCM of n elements
using System;
 
public class GFG {
     
    public static long lcm_of_array_elements(int[] element_array)
    {
        long lcm_of_array_elements = 1;
        int divisor = 2;
         
        while (true) {
             
            int counter = 0;
            bool divisible = false;
            for (int i = 0; i < element_array.Length; i++) {
 
                // lcm_of_array_elements (n1, n2, ... 0) = 0.
                // For negative number we convert into
                // positive and calculate lcm_of_array_elements.
                if (element_array[i] == 0) {
                    return 0;
                }
                else if (element_array[i] < 0) {
                    element_array[i] = element_array[i] * (-1);
                }
                if (element_array[i] == 1) {
                    counter++;
                }
 
                // Divide element_array by devisor if complete
                // division i.e. without remainder then replace
                // number with quotient; used for find next factor
                if (element_array[i] % divisor == 0) {
                    divisible = true;
                    element_array[i] = element_array[i] / divisor;
                }
            }
 
            // If divisor able to completely divide any number
            // from array multiply with lcm_of_array_elements
            // and store into lcm_of_array_elements and continue
            // to same divisor for next factor finding.
            // else increment divisor
            if (divisible) {
                lcm_of_array_elements = lcm_of_array_elements * divisor;
            }
            else {
                divisor++;
            }
 
            // Check if all element_array is 1 indicate 
            // we found all factors and terminate while loop.
            if (counter == element_array.Length) {
                return lcm_of_array_elements;
            }
        }
    }
     
    // Driver Code
    public static void Main()
    {
        int[] element_array = { 2, 7, 3, 9, 4 };
        Console.Write(lcm_of_array_elements(element_array));
    }
}
 
// This Code is contributed by nitin mittal

PHP

<?php
// PHP program to find LCM of n elements
 
// Utility function to find
// GCD of 'a' and 'b'
function gcd($a, $b)
{
    if ($b == 0)
        return $a;
    return gcd($b, $a % $b);
}
 
// Returns LCM of array elements
function findlcm($arr, $n)
{
     
    // Initialize result
    $ans = $arr[0];
 
    // ans contains LCM of 
    // arr[0], ..arr[i]
    // after i'th iteration,
    for ($i = 1; $i < $n; $i++)
        $ans = ((($arr[$i] * $ans)) /
                (gcd($arr[$i], $ans)));
 
    return $ans;
}
 
// Driver Code
$arr = array(2, 7, 3, 9, 4 );
$n = sizeof($arr);
echo findlcm($arr, $n);
 
// This code is contributed by ChitraNayal
?>

Javascript

<script>
 
// Javascript program to find LCM of n elements 
 
// Utility function to find 
// GCD of 'a' and 'b' 
function gcd(a, b) 
{ 
    if (b == 0) 
        return a; 
    return gcd(b, a % b); 
} 
 
// Returns LCM of array elements 
function findlcm(arr, n) 
{ 
    // Initialize result 
    let ans = arr[0]; 
 
    // ans contains LCM of arr[0], ..arr[i] 
    // after i'th iteration, 
    for (let i = 1; i < n; i++) 
        ans = (((arr[i] * ans)) / 
                (gcd(arr[i], ans))); 
 
    return ans; 
} 
 
// Driver Code 
  
    let arr = [ 2, 7, 3, 9, 4 ]; 
    let n = arr.length; 
    document.write(findlcm(arr, n)); 
 
// This code is contributed by Mayank Tyagi
 
</script>

Output

Below is implementation of above algorithm Recursively :

C++

#include <bits/stdc++.h>
using namespace std;
 
//recursive implementation
int LcmOfArray(vector<int> arr, int idx){
    // lcm(a,b) = (a*b/gcd(a,b))
    if (idx == arr.size()-1){
        return arr[idx];
    }
    int a = arr[idx];
    int b = LcmOfArray(arr, idx+1);
    return (a*b/__gcd(a,b)); // __gcd(a,b) is inbuilt library function
}
 
 
int main() {
     
    vector<int> arr = {1,2,8,3};
    cout << LcmOfArray(arr, 0) << "\n";
      arr = {2,7,3,9,4};
      cout << LcmOfArray(arr,0) << "\n";
    return 0;
} 

Java

import java.util.*;
class GFG
{
 
  // Recursive function to return gcd of a and b  
  static int __gcd(int a, int b)  
  {  
    return b == 0? a:__gcd(b, a % b);     
  }
 
  // recursive implementation
  static int LcmOfArray(int[] arr, int idx)
  {
 
    // lcm(a,b) = (a*b/gcd(a,b))
    if (idx == arr.length - 1){
      return arr[idx];
    }
    int a = arr[idx];
    int b = LcmOfArray(arr, idx+1);
    return (a*b/__gcd(a,b)); // 
  }
 
 
  public static void main(String[] args)
  {
 
    int[] arr = {1,2,8,3};
    System.out.print(LcmOfArray(arr, 0)+ "\n");
    int[]  arr1 = {2,7,3,9,4};
    System.out.print(LcmOfArray(arr1,0)+ "\n");
  }
} 
 
// This code is contributed by gauravrajput1 

Python3

def __gcd(a, b):
    if (a == 0):
        return b
    return __gcd(b % a, a)
 
# recursive implementation
def LcmOfArray(arr, idx):
   
    # lcm(a,b) = (a*b/gcd(a,b))
    if (idx == len(arr)-1):
        return arr[idx]
    a = arr[idx]
    b = LcmOfArray(arr, idx+1)
    return int(a*b/__gcd(a,b)) # __gcd(a,b) is inbuilt library function
 
arr = [1,2,8,3]
print(LcmOfArray(arr, 0))
arr = [2,7,3,9,4]
print(LcmOfArray(arr,0))
 
# This code is contributed by divyeshrabadiya07.

C#

using System;
class GFG {
     
    // Function to return
    // gcd of a and b
    static int __gcd(int a, int b)
    {
        if (a == 0)
            return b;
        return __gcd(b % a, a);
    }
  
    //recursive implementation
    static int LcmOfArray(int[] arr, int idx){
        // lcm(a,b) = (a*b/gcd(a,b))
        if (idx == arr.Length-1){
            return arr[idx];
        }
        int a = arr[idx];
        int b = LcmOfArray(arr, idx+1);
        return (a*b/__gcd(a,b)); // __gcd(a,b) is inbuilt library function
    }
     
  static void Main() {
    int[] arr = {1,2,8,3};
    Console.WriteLine(LcmOfArray(arr, 0));
    int[] arr1 = {2,7,3,9,4};
    Console.WriteLine(LcmOfArray(arr1,0));
  }
}

Javascript

<script>
 
    // Function to return
    // gcd of a and b
    function __gcd(a, b)
    {
        if (a == 0)
            return b;
        return __gcd(b % a, a);
    }
 
    //recursive implementation
    function LcmOfArray(arr, idx){
        // lcm(a,b) = (a*b/gcd(a,b))
        if (idx == arr.length-1){
            return arr[idx];
        }
        let a = arr[idx];
        let b = LcmOfArray(arr, idx+1);
        return (a*b/__gcd(a,b)); // __gcd(a,b) is inbuilt library function
    }
     
    let arr = [1,2,8,3];
    document.write(LcmOfArray(arr, 0) + "</br>");
    arr = [2,7,3,9,4];
    document.write(LcmOfArray(arr,0));
 
// This code is contributed by decode2207.
</script>

Output

24
252

Related Article :

This article is contributed by Madhur Modi. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

My Personal Notes

Related Articles

Table of Contents

LCM of given array elements

C++

Java

Python

C#

PHP

Javascript

C++

Java

Python3

C#

Javascript

Start Your Coding Journey Now!