**Sales by Match**

### Problem Statement :

There is a large pile of socks that must be paired by color. Given an array of integers representing the color of each sock, determine how many pairs of socks with matching colors there are. Function Description Complete the sockMerchant function in the editor below. sockMerchant has the following parameter(s): int n: the number of socks in the pile int ar[n]: the colors of each sock Returns int: the number of pairs Input Format The first line contains an integer n, the number of socks represented in ar. The second line contains n space-separated integers, ar[ i ], the colors of the socks in the pile. Constraints 1 <= n <= 100 1 <= ar[ i ] <= 100, where 0 <= i < n

### Solution :

` ````
Solution in C :
In C :
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() {
int n,i,t;
scanf("%d",&n);
int c[101];
memset(c,0,sizeof(c));
for(i=0;i<n;i++)
{
scanf("%d",&t);
c[t]++;
}
long int ans=0;
for(i=1;i<=100;i++)
{
ans+=(c[i]/2);
}
printf("%ld",ans);
return 0;
}
```

` ````
Solution in C++ :
In C++ :
#include <bits/stdc++.h>
#define FI(i,a,b) for(int i=(a);i<=(b);i++)
#define FD(i,a,b) for(int i=(a);i>=(b);i--)
#define LL long long
#define Ldouble long double
#define PI 3.1415926535897932384626
#define PII pair<int,int>
#define PLL pair<LL,LL>
#define mp make_pair
#define fi first
#define se second
using namespace std;
int n, c[105], x;
int main(){
scanf("%d", &n);
FI(i, 1, n){
scanf("%d", &x);
c[x]++;
}
int ans = 0;
FI(i, 1, 100) ans += c[i] / 2;
printf("%d\n", ans);
return 0;
}
```

` ````
Solution in Java :
In Java :
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int pairs = 0;
Set<Integer> unmatched = new HashSet<>();
for (int i = 0; i < n; ++i) {
int ci = in.nextInt();
if (unmatched.contains(ci)) {
unmatched.remove(ci);
++pairs;
}
else {
unmatched.add(ci);
}
}
System.out.println(pairs);
}
}
```

` ````
Solution in Python :
In Python3 :
#!/bin/python3
import math
import os
import random
import re
import sys
# Complete the sockMerchant function below.
from collections import Counter
def sockMerchant(n, ar):
sum = 0
for val in Counter(ar).values():
sum += val//2
return sum
if __name__ == '__main__':
fptr = open(os.environ['OUTPUT_PATH'], 'w')
n = int(input())
ar = list(map(int, input().rstrip().split()))
result = sockMerchant(n, ar)
fptr.write(str(result) + '\n')
fptr.close()
```

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