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.

1. Initialize result = 1
2. Find a common factors of two or more array elements.
3. Multiply the result by common factor and divide all the array elements by this common factor.
4. Repeat steps 2 and 3 while there is a common factor of two or more elements.
5. 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 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 indexes;         for (int j=0; j= 2)         {             // Reduce all array elements divisible             // by x.             for (int j=0; j

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 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

 # Pyhton3 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

 = 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:

420

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