# 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 arrayint 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 arrayint 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 arrayvoid findRatio(int arr[], int n){    int gcd = findGCD(arr, n);    int lcm = findLCM(arr, n);     cout << lcm / gcd << ":"         << 1 << endl;} // Driver Codeint main(){    int arr[] = { 6, 12, 36 };    int N = sizeof(arr) / sizeof(arr[0]);     findRatio(arr, N);     return 0;}

## Java

 // Java Program to implement// above approachclass GFG{      // Function to calculate and// return GCD of the given arraystatic 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 arraystatic 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 arraystatic 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 Codepublic 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 approachimport math # Function to calculate and# return GCD of the given arraydef 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 arraydef 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 arraydef findRatio(arr, n):         gcd = findGCD(arr, n)    lcm = findLCM(arr, n)         print(int(lcm / gcd), ":", "1") # Driver Codearr = [ 6, 12, 36 ]N = len(arr) findRatio(arr, N) # This code is contributed by sanjoy_62

## C#

 // C# Program to implement// above approachusing System;class GFG{      // Function to calculate and// return GCD of the given arraystatic 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 arraystatic 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 arraystatic void findRatio(int []arr, int n){    int gcd = findGCD(arr, n);    int lcm = findLCM(arr, n);     Console.Write((lcm / gcd));    Console.Write(":1");} // Driver Codepublic 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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