# Find the ratio of LCM to GCD of a given Array

Given an array arr[] of positive integers, the task is to find the ratio of LCM and GCD of the given array.
Examples:

Input: arr[] = {2, 3, 5, 9}
Output: 90:1
Explanation:
The GCD of the given array is 1 and the LCM is 90.
Therefore, the ratio is evaluated as 90:1.
Input: arr[] = {6, 12, 36}
Output: 6:1
Explanation:
The GCD of the given array is 6 and the LCM is 36.
Therefore the ratio is evaluated as 6:1.

Approach:
Follow the steps below to solve the problems:

1. First of all, we will find the GCD of the given array . For this purpose, we can use the inbuilt function for GCD provided by STL or we can use Euclidean algorithm

2. Then, we will find the LCM of the array by using the below formula:

3. At last, we will find the required ratio.

Below is the implementation of the above approach:

## C++

 // C++ Program to implement // above approach #include  using namespace std;   // Function to calculate and // return GCD of the given array int findGCD(int arr[], int n) {     // Initialise GCD     int gcd = arr[0];     for (int i = 1; i < n; i++) {         gcd = __gcd(arr[i], gcd);           // Once GCD is 1, it         // will always be 1 with         // all other elements         if (gcd == 1) {             return 1;         }     }       // Return GCD     return gcd; }   // Function to calculate and // return LCM of the given array int findLCM(int arr[], int n) {     // Initialise LCM     int lcm = arr[0];       // LCM of two numbers is     // evaluated as [(a*b)/gcd(a, b)]     for (int i = 1; i < n; i++) {         lcm = (((arr[i] * lcm))                / (__gcd(arr[i], lcm)));     }       // Return LCM     return lcm; }   // Function to print the ratio // of LCM to GCD of the given array void findRatio(int arr[], int n) {     int gcd = findGCD(arr, n);     int lcm = findLCM(arr, n);       cout << lcm / gcd << ":"          << 1 << endl; }   // Driver Code int main() {     int arr[] = { 6, 12, 36 };     int N = sizeof(arr) / sizeof(arr[0]);       findRatio(arr, N);       return 0; }

## Java

 // Java Program to implement // above approach class GFG{        // Function to calculate and // return GCD of the given array static int __gcd(int a, int b)  {      if (b == 0)          return a;      return __gcd(b, a % b);  }   static int findGCD(int arr[], int n) {     // Initialise GCD     int gcd = arr[0];     for (int i = 1; i < n; i++)      {         gcd = __gcd(arr[i], gcd);           // Once GCD is 1, it         // will always be 1 with         // all other elements         if (gcd == 1)          {             return 1;         }     }       // Return GCD     return gcd; }   // Function to calculate and // return LCM of the given array static int findLCM(int arr[], int n) {     // Initialise LCM     int lcm = arr[0];       // LCM of two numbers is     // evaluated as [(a*b)/gcd(a, b)]     for (int i = 1; i < n; i++)      {         lcm = (((arr[i] * lcm)) /            (__gcd(arr[i], lcm)));     }       // Return LCM     return lcm; }   // Function to print the ratio // of LCM to GCD of the given array static void findRatio(int arr[], int n) {     int gcd = findGCD(arr, n);     int lcm = findLCM(arr, n);       System.out.print((lcm / gcd));     System.out.print(":1"); }   // Driver Code public static void main (String[] args)  {     int arr[] = new int[]{ 6, 12, 36 };     int N = 3;       findRatio(arr, N); } }   // This code is contributed by Ritik Bansal

## Python3

 # Python3 program to implement # above approach import math   # Function to calculate and # return GCD of the given array def findGCD(arr, n):           # Initialise GCD     gcd = arr[0]       for i in range(1, n):         gcd = int(math.gcd(arr[i], gcd))                   # Once GCD is 1, it         # will always be 1 with         # all other elements         if (gcd == 1):             return 1                   # Return GCD     return gcd   # Function to calculate and # return LCM of the given array def findLCM(arr, n):           # Initialise LCM     lcm = arr[0]       # LCM of two numbers is     # evaluated as [(a*b)/gcd(a, b)]     for i in range(1, n):         lcm = int((((arr[i] * lcm)) /            (math.gcd(arr[i], lcm))))       # Return LCM     return lcm   # Function to print the ratio # of LCM to GCD of the given array def findRatio(arr, n):           gcd = findGCD(arr, n)     lcm = findLCM(arr, n)           print(int(lcm / gcd), ":", "1")   # Driver Code arr = [ 6, 12, 36 ] N = len(arr)   findRatio(arr, N)   # This code is contributed by sanjoy_62

## C#

 // C# Program to implement // above approach using System; class GFG{        // Function to calculate and // return GCD of the given array static int __gcd(int a, int b)  {      if (b == 0)          return a;      return __gcd(b, a % b);  }   static int findGCD(int []arr, int n) {     // Initialise GCD     int gcd = arr[0];     for (int i = 1; i < n; i++)      {         gcd = __gcd(arr[i], gcd);           // Once GCD is 1, it         // will always be 1 with         // all other elements         if (gcd == 1)          {             return 1;         }     }       // Return GCD     return gcd; }   // Function to calculate and // return LCM of the given array static int findLCM(int []arr, int n) {     // Initialise LCM     int lcm = arr[0];       // LCM of two numbers is     // evaluated as [(a*b)/gcd(a, b)]     for (int i = 1; i < n; i++)      {         lcm = (((arr[i] * lcm)) /            (__gcd(arr[i], lcm)));     }       // Return LCM     return lcm; }   // Function to print the ratio // of LCM to GCD of the given array static void findRatio(int []arr, int n) {     int gcd = findGCD(arr, n);     int lcm = findLCM(arr, n);       Console.Write((lcm / gcd));     Console.Write(":1"); }   // Driver Code public static void Main()  {     int []arr = new int[]{ 6, 12, 36 };     int N = 3;       findRatio(arr, N); } }   // This code is contributed by Code_Mech

## Javascript

 

Output:

6:1

Time Complexity: O(N * logN)
Auxiliary Space: O(1)

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