Count equal element pairs in the given array

Given an array arr[] of N integers representing the lengths of the gloves, the task is to count the maximum possible pairs of gloves from the given array. Note that a glove can only pair with a same sized glove and it can only be part of a single pair.

Examples:

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

Simple Approach: Sort the given array so that all the equal elements are adjacent to each other. Now, traverse the array and for every element if it equal to the element next to it then it is a valid pair and skip these two elements. Else the current element doesn’t make a valid pair with any other element and hence only skip the current element.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
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]);
  
    cout << cntgloves(arr, n);
  
    return 0;
}

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Java

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// 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)
{
    // To store the required count
    int count = 0;
  
    // Sort the original array
    Arrays.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;
  
    System.out.println(cntgloves(arr, n));
}
}
  
// This code is contributed by PrinciRaj1992

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Python3

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# 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); 
  
    print(cntgloves(arr, n)); 
  
    # This code is contributed by AnkitRai01

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C#

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// 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;
  
    Console.WriteLine(cntgloves(arr, n));
}
}
  
// This code is contributed by Princi Singh

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Output:

2

Efficient Approach
1) Create an empty hash table (unordered_map in C++, HashMap in Java, Dictionary in Python)
2) Store frequencies of all elements.
3) Traverse through the hash table. For every element, find its frequency. Increment the result by frequency/2 for every element,



My Personal Notes arrow_drop_up


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