查找给定 Array 中出现次数最多和出现次数最少的元素的 GCD

给定一个大小为n的数组arr[] ，任务是找到给定数组中最高和最低频率元素的 GCD。

例子：

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

方法：这个想法是计算所有元素的频率并将它们存储在一个向量中，然后找到出现次数最多和出现次数最少的元素的 gcd。

下面是上述方法的实现：

C++14

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find frequency and find gcd
int find_gcd(vector v, int n)
{
    map 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;
}
 
// Drive Code
int main()
{
    vector v = { 2, 2, 4, 4, 5, 5, 6, 6, 6, 6 };
    int n = v.size();
    cout << find_gcd(v, n) << endl;
 
    vector 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 mp = new HashMap();
 
    // 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 it : mp.entrySet()) {
        mini = mp.get(mini) < it.getValue()
                   ? it.getKey()
                   : mini;
    }
    for (Map.Entry 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
 
# Drive 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 mp
            = new Dictionary();
 
        // 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 it in mp)
        {
            mini = mp[mini] < it.Value ? it.Key : mini;
        }
        foreach(KeyValuePair 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

       // 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
   

输出：

2
1

时间复杂度： O(n*log(n))
辅助空间： O(n)