Shirt - Matching Pairs

 Shirt - Matching Pairs

PROBLEM STATEMENT :

A shop to increase sales during a festival has an offer that a customer will get a discount if the customer buys shirts having same size in pairs. Any customer who buys will choose N shirts and the size of the shirt is denoted by S(i) where 1 <= i <=N. Two shirts S(i) and S(j) are matching and form a pair only if S(i) = S(j). 

The program must print the number of pairs eligible for the discount. 

Input Format: 

The first line will contain the value of N 

The second line will contain the the size of N shirts S(1) to S(N) with each size separated by a space.

Output Format: 

The first line will contain the number of matching pairs eligible for the discount. 

Constraints: 

2 <= N <= 100 

Example Input/Output 1: 
Input: () 
9 
10 20 20 10 10 30 44 10 20 
Output: 
3 

Explanation: 

The matching pairs are (10,10) (20,20) (10,10). 

Example Input/Output 2: 
Input: () 
6 
42 44 40 42 44 42 
Output: 
2

---

SOLUTION :

C (Programming Language)

      
#include<stdio.h>
#include<stdlib.h>

int main()
{
    int n,i,a,cnt;
    scanf("%d",&n);
    int hash[1001];
    for(i=0;i<n;++i)
    {
        scanf("%d",&a);
        hash[a]++;
        if(hash[a]%2==0) cnt++;
    }
    printf("%d",cnt);
}
 

C++ (CPP)

       
#include <iostream>
#include<map>
using namespace std;

int main(int argc, char** argv)
{
    int n,a,i,cnt=0;
    cin>>n;
    map<int,int> hm;
    for(i=0;i<n;++i)
    {
        cin>>a;
        hm[a]++;
        if(hm[a]%2==0) cnt+=1;
    }
    cout<<cnt;
}
 

JAVA

       
import java.util.*;
public class Hello {

    public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		int n=sc.nextInt(),i,a,cnt=0;
		HashMap<Integer,Integer> hm=new HashMap<Integer,Integer>();
		for(i=0;i<n;++i)
		{
		    a=sc.nextInt();
		    if(!hm.containsKey(a)) hm.put(a,1);
		    else hm.put(a,hm.get(a)+1);
		    if((hm.get(a))%2==0) cnt+=1;
		}
		System.out.println(cnt);
	}
}
 

PYTHON

       
n=int(input())
l=list(map(int,input().split()))
done=[];res=0
for i in range(n):
    if l[i] not in done:
        res+=(l.count(l[i]))//2
        done.append(l[i])
print(res)
 

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