# 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.

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

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;      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;         // 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);         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;      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;         // 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         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         # 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;      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;         // 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

Output:

6:1


Time Complexity: O(N * logN)
Auxiliary Space: O(1) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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