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# Count equal element pairs in the given array

• Last Updated : 25 May, 2021

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:
(arr, arr) and (arr, arr) are the only possible pairs.

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

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++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);` `    ``// 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`

## Javascript

 ``
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,

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