✨ อัปเดตล่าสุด: web//https://www.geeksforgeeks.org/finding lcm two 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:

This article is contributed by Aditya Kumar. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@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.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes

Nov	DEC	Jan
	02
2019	2020	2021

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

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

C++

Java

Python3

C#

PHP

Recommended Posts: