# 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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## Find Merge Point of Two Lists

This challenge is part of a tutorial track by MyCodeSchool Given pointers to the head nodes of 2 linked lists that merge together at some point, find the node where the two lists merge. The merge point is where both lists point to the same node, i.e. they reference the same memory location. It is guaranteed that the two head nodes will be different, and neither will be NULL. If the lists share

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## Reverse a doubly linked list

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## Tree: Preorder Traversal

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