计算给定数组中的相等元素对

给定一个代表手套长度的N个整数的数组arr [] ，任务是从给定的数组中计算最大可能的手套对。请注意，手套只能与相同尺寸的手套配对，并且只能是一对手套的一部分。

例子：

Input: arr[] = {6, 5, 2, 3, 5, 2, 2, 1} 
Output: 2 
(arr[1], arr[4]) and (arr[2], arr[5]) are the only possible pairs.

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

简单方法：对给定的数组进行排序，以使所有相等的元素彼此相邻。现在，遍历数组，对于每个元素，如果它等于其旁边的元素，则为有效对，并跳过这两个元素。否则，当前元素不会与任何其他元素形成有效对，因此只能跳过当前元素。

下面是上述方法的实现：

C++14

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the maximum
// possible pairs of gloves
int cntgloves(int arr[], int n)
{
    // To store the required count
    int count = 0;
 
    // Sort the original array
    sort(arr, arr + n);
 
    for (int i = 0; i < n - 1;) {
 
        // A valid pair is found
        if (arr[i] == arr[i + 1]) {
            count++;
 
            // Skip the elements of
            // the current pair
            i = i + 2;
        }
 
        // Current elements doesn't make
        // a valid pair with any other element
        else {
            i++;
        }
    }
    return count;
}
 
// Driver code
int main()
{
    int arr[] = { 6, 5, 2, 3, 5, 2, 2, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    // Function call
    cout << cntgloves(arr, n);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG {
 
    // Function to return the maximum
    // possible pairs of gloves
    static int cntgloves(int arr[], int n)
    {
         
        // Sort the original array
        Arrays.sort(arr);
        int res = 0;
        int i = 0;
 
        while (i < n) {
             
            // take first number
            int number = arr[i];
            int count = 1;
            i++;
 
            // Count all duplicates
            while (i < n && arr[i] == number) {
                count++;
                i++;
            }
             
            // If we spotted number just 2
            // times, increment
            // result
            if (count >= 2) {
                res = res + count / 2;
            }
        }
        return res;
    }
 
    // Driver code
    public static void main(String[] args)
    {
 
        int arr[] = {6, 5, 2, 3, 5, 2, 2, 1};
        int n = arr.length;
 
        // Function call
        System.out.println(cntgloves(arr, n));
    }
}
 
// This code is contributed by Lakhan murmu

Python3

# Python3 implementation of the approach
 
# Function to return the maximum
# possible pairs of gloves
 
 
def cntgloves(arr, n):
 
    # To store the required count
    count = 0
 
    # Sort the original array
    arr.sort()
    i = 0
    while i < (n-1):
 
        # A valid pair is found
        if (arr[i] == arr[i + 1]):
            count += 1
 
            # Skip the elements of
            # the current pair
            i = i + 2
 
        # Current elements doesn't make
        # a valid pair with any other element
        else:
            i += 1
 
    return count
 
 
# Driver code
if __name__ == "__main__":
 
    arr = [6, 5, 2, 3, 5, 2, 2, 1]
    n = len(arr)
 
    # Function call
    print(cntgloves(arr, n))
 
    # This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
 
class GFG {
 
    // Function to return the maximum
    // possible pairs of gloves
    static int cntgloves(int[] arr, int n)
    {
        // To store the required count
        int count = 0;
 
        // Sort the original array
        Array.Sort(arr);
 
        for (int i = 0; i < n - 1;) {
 
            // A valid pair is found
            if (arr[i] == arr[i + 1]) {
                count++;
 
                // Skip the elements of
                // the current pair
                i = i + 2;
            }
 
            // Current elements doesn't make
            // a valid pair with any other element
            else {
                i++;
            }
        }
        return count;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int[] arr = { 6, 5, 2, 3, 5, 2, 2, 1 };
        int n = arr.Length;
 
        // Function call
        Console.WriteLine(cntgloves(arr, n));
    }
}
 
// This code is contributed by Princi Singh

输出：

2

高效方法
1)创建一个空的哈希表(在C++中为unordered_map，在JavaHashMap，在PythonDictionary)
2)存储所有元素的频率。
3)遍历哈希表。对于每个元素，找到其频率。对于每个元素，将结果按频率/ 2递增，