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 = 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 in 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; }

Output:

420

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.

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