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

22 Apr, 2021
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
 
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 = 1Find 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.
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++
Java
Python3
C#
PHP
Javascript
// 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 codeint 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;}
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 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 codearr = [1, 2, 3, 4, 5, 10, 20, 35];n = len(arr);print(LCM(arr, n)); # This code is contributed by chandan_jnu
// 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 codepublic 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 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?>
<script>// Javascript 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[]    var max_num = 0;    for (var i = 0; i < n; i++)        if (max_num < arr[i])            max_num = arr[i];     // Initialize result    var res = 1;     // Find all factors that are present in    // two or more array elements.    var x = 2;  // Current factor.    while (x <= max_num)    {        // To store indexes of all array        // elements that are divisible by x.        var indexes = [];        for (var j = 0; j < n; j++)            if (arr[j] % x == 0)                indexes.push(j);         // If there are 2 or more array elements        // that are divisible by x.        if (indexes.length >= 2)        {            // Reduce all array elements divisible            // by x.            for (var j = 0; j < indexes.length; j++)                arr[indexes[j]] = arr[indexes[j]]/x;             res = res * x;        }        else            x++;    }     // Then multiply all reduced array elements    for (var i = 0; i < n; i++)        res = res*arr[i];     return res;} // Driver codevar arr = [1, 2, 3, 4, 5, 10, 20, 35];var n = arr.length;document.write( LCM(arr, n) + "<br>"); // This code is contributed by rrrtnx.</script>
Output:  
420
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