Find GCD of most occurring and least occurring elements of given Array

Given an array arr[] of size n, The task is to find the GCD of the highest and lowest frequency element in the given array.

Examples:

Input: arr[] = {2, 2, 4, 4, 5, 5, 6, 6, 6, 6}
Output: 2
Explanation: The frequency of the elements in the above array is 
freq(2) = 2, 
freq(4) = 2, 
freq(5) = 2, 
freq(6) = 4, 
The minimum frequency is 2 (of elements 2, 4, and 5). So 2 will be picked as the least among 2, 4, and 5. 
The largest frequency is 4 (of element 6). 
Hence GCD of 2 and 6 = gcd(2, 6) is 2.

Input: arr[] = {3, 2, 2, 44, 44, 44, 44}
Output: 1

Approach: The idea is to count the frequencies of all elements and store them in a vector and then find the gcd of the highest occurring and the least occurring elements.

Algorithm:

• Define a function find_gcd that takes a vector of integers v and an integer n representing the size of the vector.
• Declare a map data structure mp that will store the frequency of each element in the vector.
• Iterate through each element in the vector v using a for loop with index i, and update the frequency of each element in the map data structure mp.
• Initialize two variables mini and maxi to the first element in the vector v.
• Iterate through each element in the map data structure mp using a range-based for loop with auto keyword.
• For each element in the map data structure mp, check if its frequency is less than the frequency of the current minimum element mini. If it is, update the value of mini to the current element. Otherwise, keep the current mini value.
• For each element in the map data structure mp, check if its frequency is greater than the frequency of the current maximum element maxi. If it is, update the value of maxi to the current element. Otherwise, keep the current maxi value.
• Use the __gcd function from the numeric library to find the GCD of the mini and maxi values, and store the result in the res variable.
• Return the res variable.
• In the main function, initialize two vectors v and v1 with integer values.
• Call the find_gcd function with v and v1, and print the returned value.

Below is the implementation of the above approach:

C++14

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find frequency and find gcd
int find_gcd(vector<int> v, int n)
{
    map<int, int> mp;

    // Update the frequency
    for (int i = 0; i < n; i++) {
        mp[v[i]]++;
    }

    int mini = v[0], maxi = v[0];

    for (auto it : mp) {
        mini = mp[mini] < it.second
                   ? it.first
                   : mini;
    }
    for (auto it : mp) {
        maxi = mp[maxi] > it.second
                   ? it.first
                   : maxi;
    }

    // Find gcd
    int res = __gcd(mini, maxi);
    return res;
}

// Driver Code
int main()
{
    vector<int> v = { 2, 2, 4, 4, 5, 5, 6, 6, 6, 6 };
    int n = v.size();
    cout << find_gcd(v, n) << endl;

    vector<int> v1 = { 3, 2, 2, 44, 44, 44, 44 };
    cout << find_gcd(v1, v1.size()) << endl;

    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{

// Function to find frequency and find gcd
static int find_gcd(int []v, int n)
{
    HashMap<Integer,Integer> mp = new HashMap<Integer,Integer>();

    // Update the frequency
    for (int i = 0; i < n; i++) {
         if(mp.containsKey(v[i])){
                mp.put(v[i], mp.get(v[i])+1);
            }
            else{
                mp.put(v[i], 1);
            }
    }

    int mini = v[0], maxi = v[0];

    for (Map.Entry<Integer,Integer> it : mp.entrySet()) {
        mini = mp.get(mini) < it.getValue()
                   ? it.getKey()
                   : mini;
    }
    for (Map.Entry<Integer,Integer> it : mp.entrySet()) {
        maxi = mp.get(maxi) > it.getValue()
                   ? it.getKey()
                   : maxi;
    }

    // Find gcd
    int res = __gcd(mini, maxi);
    return res;
}
static int __gcd(int a, int b)
{
    return b == 0? a:__gcd(b, a % b);
}

// Drive Code
public static void main(String[] args)
{
    int [] v = { 2, 2, 4, 4, 5, 5, 6, 6, 6, 6 };
    int n = v.length;
    System.out.print(find_gcd(v, n) +"\n");

    int [] v1 = { 3, 2, 2, 44, 44, 44, 44 };
    System.out.print(find_gcd(v1, v1.length) +"\n");
}
}

// This code is contributed by shikhasingrajput

Python3

# python program for the above approach
import math

# Function to find frequency and find gcd
def find_gcd(v, n):

    mp = {}

    # Update the frequency
    for i in range(0, n):
        if v[i] in mp:
            mp[v[i]] += 1
        else:
            mp[v[i]] = 1

    mini = v[0]
    maxi = v[0]

    for it in mp:
        if mini in mp and mp[mini] < mp[it]:
            mini = it

    for it in mp:
        if mp[maxi] > mp[it]:
            maxi = it

        # Find gcd
    res = math.gcd(mini, maxi)
    return res

# Driver Code
if __name__ == "__main__":

    v = [2, 2, 4, 4, 5, 5, 6, 6, 6, 6]
    n = len(v)
    print(find_gcd(v, n))

    v1 = [3, 2, 2, 44, 44, 44, 44]
    print(find_gcd(v1, len(v1)))

    # This code is contributed by rakeshsahni

C#

// C# program for the above approach
using System;
using System.Collections.Generic;

class GFG {

    static int __gcd(int a, int b)
    {
        return b == 0 ? a : __gcd(b, a % b);
    }

    // Function to find frequency and find gcd
    static int find_gcd(int[] v, int n)
    {
        Dictionary<int, int> mp
            = new Dictionary<int, int>();

        // Update the frequency
        for (int i = 0; i < n; i++) {
            if (mp.ContainsKey(v[i])) {
                mp[v[i]] += 1;
            }
            else {
                mp[v[i]] = 1;
            }
        }

        int mini = v[0], maxi = v[0];

        foreach(KeyValuePair<int, int> it in mp)
        {
            mini = mp[mini] < it.Value ? it.Key : mini;
        }
        foreach(KeyValuePair<int, int> it in mp)
        {
            maxi = mp[maxi] > it.Value ? it.Key : maxi;
        }

        // Find gcd
        int res = __gcd(mini, maxi);
        return res;
    }

    // Drive Code
    public static void Main()
    {
        int[] v = { 2, 2, 4, 4, 5, 5, 6, 6, 6, 6 };
        int n = v.Length;
        Console.WriteLine(find_gcd(v, n));

        int[] v1 = { 3, 2, 2, 44, 44, 44, 44 };
        Console.WriteLine(find_gcd(v1, v1.Length));
    }
}

// This code is contributed by ukasp.

Javascript

<script>
       // JavaScript code for the above approach
       function __gcd(a, b) {
           // Everything divides 0
           if (a == 0)
               return b;
           if (b == 0)
               return a;

           // base case
           if (a == b)
               return a;

           // a is greater
           if (a > b)
               return __gcd(a - b, b);
           return __gcd(a, b - a);
       }
       // Function to find frequency and find gcd
       function find_gcd(v, n) {
           let mp = new Map();

           // Update the frequency
           for (let i = 0; i < n; i++) {

               if (!mp.has(v[i]))
                   mp.set(v[i], 1);
               else
                   mp.set(v[i], mp.get(v[i]) + 1)
           }

           let mini = v[0], maxi = v[0];

           for (let [key, val] of mp) {
               mini = mp.get(mini) < val
                   ? key
                   : mini;
           }
           for (let [key, val] of mp) {
               maxi = mp.get(maxi) > val
                   ? key
                   : maxi;
           }

           // Find gcd
           let res = __gcd(mini, maxi);
           return res;
       }

       // Drive Code
       let v = [2, 2, 4, 4, 5, 5, 6, 6, 6, 6];
       let n = v.length;
       document.write(find_gcd(v, n) + '<br>');

       let v1 = [3, 2, 2, 44, 44, 44, 44];
       document.write(find_gcd(v1, v1.length) + '<br>');

 // This code is contributed by Potta Lokesh
   </script>

Output:

2
1

Time Complexity: O(n*log(n))
Auxiliary Space: O(n)