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Find the ratio of LCM to GCD of a given Array
  • Last Updated : 14 Apr, 2021
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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: 
    LCM(a, b)=\frac{a*b}{gcd(a, b)}
     
  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 <bits/stdc++.h>
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




<script>
// javascript Program to implement
// above approach   
 
    // Function to calculate and
    // return GCD of the given array
    function __gcd(a , b)
    {
        if (b == 0)
            return a;
        return __gcd(b, a % b);
    }
 
    function findGCD(arr, n)
    {
     
        // Initialise GCD
        var gcd = arr[0];
        for (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
    function findLCM(arr, n)
    {
     
        // Initialise LCM
        var lcm = arr[0];
 
        // LCM of two numbers is
        // evaluated as [(a*b)/gcd(a, b)]
        for (i = 1; i < n; i++)
        {
            lcm = (((arr[i] * lcm)) / (__gcd(arr[i], lcm)));
        }
 
        // Return LCM
        return lcm;
    }
 
    // Function to prvar the ratio
    // of LCM to GCD of the given array
    function findRatio(arr , n) {
        var gcd = findGCD(arr, n);
        var lcm = findLCM(arr, n);
 
        document.write((lcm / gcd));
        document.write(":1");
    }
 
    // Driver Code
        var arr = [ 6, 12, 36 ];
        var N = 3;
 
        findRatio(arr, N);
 
// This code is contributed by todaysgaurav.
</script>
Output: 
6:1

 

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

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