**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. Example n = 7 ar = [1, 2, 1, 2, 1, 3, 2] There is one pair of color 1 and one of color 2. There are three odd socks left, one of each color. The number of pairs is 2. 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 :
python 3 :
from collections import Counter
n = int(input())
c = Counter(map(int, input().split()))
ans = 0
for x in c:
ans += c[x] // 2
print(ans)
Java :
import java.io.*;
import java.util.*;
public class Solution {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n=in.nextInt();
int[] hash = new int[300];
for(int i=0; i<n; i++){
hash[in.nextInt()]++;
}
long ans=0;
for(int p: hash){
ans+=p/2;
}
System.out.println(ans);
}
}
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;
}
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;
}
```

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