Finding LCM of more than two (or array) numbers without using GCD

Given an array of positive integers, find LCM of the elements present in array.

Examples:

Input : arr[] = {1, 2, 3, 4, 28}
Output : 84

Input  : arr[] = {4, 6, 12, 24, 30}
Output : 120

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

We have discussed LCM of array using GCD.

In this post a different approach is discussed that doesn’t require computation of GCD. Below are steps.

Initialize result = 1
Find a common factors of two or more array elements.
Multiply the result by common factor and divide all the array elements by this common factor.
Repeat steps 2 and 3 while there is a common factor of two or more elements.
Multiply the result by reduced (or divided) array elements.

Illustration :

Let we have to find the LCM of 
arr[] = {1, 2, 3, 4, 28}

We initialize result = 1.

2 is a common factor that appears in
two or more elements. We divide all
multiples by two and multiply result
with 2.
arr[] = {1, 1, 3, 2, 14}
result = 2

2 is again a common factor that appears 
in two or more elements. We divide all
multiples by two and multiply result
with 2.
arr[] = {1, 1, 3, 1, 7}
result = 4

Now there is no common factor that appears
in two or more array elements. We multiply
all modified array elements with result, we
get.
result = 4 * 1 * 1 * 3 * 1 * 7
       = 84

Below is the implementation of above algorithm.

C++

// C++ program to find LCM of array without 
// using GCD. 
#include<bits/stdc++.h> 
using namespace std; 
  
// Returns LCM of arr[0..n-1] 
unsigned long long int LCM(int arr[], int n) 
{ 
    // Find the maximum value in arr[] 
    int max_num = 0; 
    for (int i=0; i<n; i++) 
        if (max_num < arr[i]) 
            max_num = arr[i]; 
  
    // Initialize result 
    unsigned long long int res = 1; 
  
    // Find all factors that are present in 
    // two or more array elements. 
    int x = 2;  // Current factor. 
    while (x <= max_num) 
    { 
        // To store indexes of all array 
        // elements that are divisible by x. 
        vector<int> indexes; 
        for (int j=0; j<n; j++) 
            if (arr[j]%x == 0) 
                indexes.push_back(j); 
  
        // If there are 2 or more array elements 
        // that are divisible by x. 
        if (indexes.size() >= 2) 
        { 
            // Reduce all array elements divisible 
            // by x. 
            for (int j=0; j<indexes.size(); j++) 
                arr[indexes[j]] = arr[indexes[j]]/x; 
  
            res = res * x; 
        } 
        else
            x++; 
    } 
  
    // Then multiply all reduced array elements 
    for (int i=0; i<n; i++) 
        res = res*arr[i]; 
  
    return res; 
} 
  
// Driver code 
int main() 
{ 
    int arr[] = {1, 2, 3, 4, 5, 10, 20, 35}; 
    int n = sizeof(arr)/sizeof(arr[0]); 
    cout << LCM(arr, n) << "\n"; 
    return 0; 
} 

Java

import java.util.Vector; 
  
// Java program to find LCM of array without  
// using GCD. 
class GFG { 
  
// Returns LCM of arr[0..n-1]  
    static long LCM(int arr[], int n) { 
        // Find the maximum value in arr[]  
        int max_num = 0; 
        for (int i = 0; i < n; i++) { 
            if (max_num < arr[i]) { 
                max_num = arr[i]; 
            } 
        } 
  
        // Initialize result  
        long res = 1; 
  
        // Find all factors that are present in  
        // two or more array elements.  
        int x = 2; // Current factor.  
        while (x <= max_num) { 
            // To store indexes of all array  
            // elements that are divisible by x.  
            Vector<Integer> indexes = new Vector<>(); 
            for (int j = 0; j < n; j++) { 
                if (arr[j] % x == 0) { 
                    indexes.add(indexes.size(), j); 
                } 
            } 
  
            // If there are 2 or more array elements  
            // that are divisible by x.  
            if (indexes.size() >= 2) { 
                // Reduce all array elements divisible  
                // by x.  
                for (int j = 0; j < indexes.size(); j++) { 
                    arr[indexes.get(j)] = arr[indexes.get(j)] / x; 
                } 
  
                res = res * x; 
            } else { 
                x++; 
            } 
        } 
  
        // Then multiply all reduced array elements  
        for (int i = 0; i < n; i++) { 
            res = res * arr[i]; 
        } 
  
        return res; 
    } 
  
// Driver code  
    public static void main(String[] args) { 
        int arr[] = {1, 2, 3, 4, 5, 10, 20, 35}; 
        int n = arr.length; 
        System.out.println(LCM(arr, n)); 
    } 
} 

Python3

# Python3 program to find LCM of array  
# without using GCD. 
  
# Returns LCM of arr[0..n-1] 
def LCM(arr, n): 
      
    # Find the maximum value in arr[] 
    max_num = 0; 
    for i in range(n): 
        if (max_num < arr[i]): 
            max_num = arr[i]; 
  
    # Initialize result 
    res = 1; 
  
    # Find all factors that are present  
    # in two or more array elements. 
    x = 2; # Current factor. 
    while (x <= max_num): 
          
        # To store indexes of all array 
        # elements that are divisible by x. 
        indexes = []; 
        for j in range(n): 
            if (arr[j] % x == 0): 
                indexes.append(j); 
  
        # If there are 2 or more array  
        # elements that are divisible by x. 
        if (len(indexes) >= 2): 
              
            # Reduce all array elements  
            # divisible by x. 
            for j in range(len(indexes)): 
                arr[indexes[j]] = int(arr[indexes[j]] / x); 
  
            res = res * x; 
        else: 
            x += 1; 
  
    # Then multiply all reduced  
    # array elements 
    for i in range(n): 
        res = res * arr[i]; 
  
    return res; 
  
# Driver code 
arr = [1, 2, 3, 4, 5, 10, 20, 35]; 
n = len(arr); 
print(LCM(arr, n)); 
  
# This code is contributed by chandan_jnu 

C#

// C# program to find LCM of array  
// without using GCD.  
using System; 
using System.Collections; 
class GFG 
{  
  
// Returns LCM of arr[0..n-1]  
static long LCM(int []arr, int n)  
{  
    // Find the maximum value in arr[]  
    int max_num = 0;  
    for (int i = 0; i < n; i++) 
    {  
        if (max_num < arr[i]) 
        {  
            max_num = arr[i];  
        }  
    }  
  
    // Initialize result  
    long res = 1;  
  
    // Find all factors that are present  
    // in two or more array elements.  
    int x = 2; // Current factor.  
    while (x <= max_num) 
    {  
        // To store indexes of all array  
        // elements that are divisible by x.  
        ArrayList indexes = new ArrayList();  
        for (int j = 0; j < n; j++) 
        {  
            if (arr[j] % x == 0) 
            {  
                indexes.Add(j);  
            }  
        }  
  
        // If there are 2 or more array elements  
        // that are divisible by x.  
        if (indexes.Count >= 2)  
        {  
            // Reduce all array elements divisible  
            // by x.  
            for (int j = 0; j < indexes.Count; j++)  
            {  
                arr[(int)indexes[j]] = arr[(int)indexes[j]] / x;  
            }  
  
            res = res * x;  
        } else 
        {  
            x++;  
        }  
    }  
  
    // Then multiply all reduced  
    // array elements  
    for (int i = 0; i < n; i++)  
    {  
        res = res * arr[i];  
    }  
  
    return res;  
}  
  
// Driver code  
public static void Main() 
{  
    int []arr = {1, 2, 3, 4, 5, 10, 20, 35};  
    int n = arr.Length;  
    Console.WriteLine(LCM(arr, n));  
}  
}  
  
// This code is contributed by mits 

PHP

<?php 
// PHP program to find LCM of array  
// without using GCD. 
  
// Returns LCM of arr[0..n-1] 
function LCM($arr, $n) 
{ 
    // Find the maximum value in arr[] 
    $max_num = 0; 
    for ($i = 0; $i < $n; $i++) 
        if ($max_num < $arr[$i]) 
            $max_num = $arr[$i]; 
  
    // Initialize result 
    $res = 1; 
  
    // Find all factors that are present  
    // in two or more array elements. 
    $x = 2; // Current factor. 
    while ($x <= $max_num) 
    { 
        // To store indexes of all array 
        // elements that are divisible by x. 
        $indexes = array(); 
        for ($j = 0; $j < $n; $j++) 
            if ($arr[$j] % $x == 0) 
                array_push($indexes, $j); 
  
        // If there are 2 or more array  
        // elements that are divisible by x. 
        if (count($indexes) >= 2) 
        { 
            // Reduce all array elements  
            // divisible by x. 
            for ($j = 0; $j < count($indexes); $j++) 
                $arr[$indexes[$j]] = (int)($arr[$indexes[$j]] / $x); 
  
            $res = $res * $x; 
        } 
        else
            $x++; 
    } 
  
    // Then multiply all reduced  
    // array elements 
    for ($i = 0; $i < $n; $i++) 
        $res = $res * $arr[$i]; 
  
    return $res; 
} 
  
// Driver code 
$arr = array(1, 2, 3, 4, 5, 10, 20, 35); 
$n = count($arr); 
echo LCM($arr, $n) . "\n"; 
  
// This code is contributed by chandan_jnu 
?> 

Output:

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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