Beautiful Pairs

                            Solution in C :

In  C :





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

int main() {

    int n;
    scanf("%d",&n);
    int a[n],b[n],i,j,count=0;
    for(i=0;i<=n-1;i++)
        {
        scanf("%d",&a[i]);
    }
    for(i=0;i<=n-1;i++)
        {
        scanf("%d",&b[i]);
    }
    for(i=0;i<=n-1;i++)
        {
        for(j=0;j<=n-1;j++)
            {
            if(b[j]!=-1)
                {
            if(a[i]==b[j])
                {
                count++;
                b[j]=-1;
                break;
            }
            }
        }
    }
    if(count==n)
        printf("%d",count-1);
    else
    printf("%d",count+1);
    return 0;
}
                        

                        Solution in C++ :

In  C++  :







#include <bits/stdc++.h>
using namespace std;

int n;
int cntA[1001];
int cntB[1001];

int MAIN()
{
	memset(cntA, 0, sizeof(cntA));
	memset(cntB, 0, sizeof(cntB));
	cin >> n;
	for(int i = 1; i <= n; i++)
	{
		int t;
		cin >> t;
		cntA[t] ++;
	}
	for(int i = 1; i <= n; i++)
	{
		int t;
		cin >> t;
		cntB[t] ++;
	}
	int ans = 0;
	for(int i = 1; i <= 1000; i++)
		ans += min(cntA[i], cntB[i]);
	if(ans == n)
		ans --;
	else
		ans ++;
	cout << ans << endl;
	
	return 0;
}

int main()
{
	int start = clock();
	#ifdef LOCAL_TEST
		freopen("in.txt", "r", stdin);
		freopen("out.txt", "w", stdout);
	#endif
	ios :: sync_with_stdio(false);
	cout << fixed << setprecision(16);
	int ret = MAIN();
	#ifdef LOCAL_TEST
		cout << "[Finished in " << clock() - start << " ms]" << endl;
	#endif
	return ret;
}
                    

                        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 s=new Scanner(System.in);
        int len=s.nextInt();
        HashMap<Integer,Integer> hmap=new HashMap<Integer,Integer>();
        for(int i=0;i<len;i++)
            {
            int current=s.nextInt();
            if(!hmap.containsKey(current))
                {
                
                hmap.put(current,1);
            }
            else
                {
                int temp=hmap.get(current);
                hmap.put(current,++temp);
            }
        }
        int counter=0;
        for(int i=0;i<len;i++)
            {
            int current=s.nextInt();
            if(hmap.containsKey(current))
            {
                int temp=hmap.get(current);
                if(temp>0)
                {
                    hmap.put(current,--temp);
                    counter++;
                }
                else
                    {
                    hmap.remove(current);
                }
                
            }
        }
        if(counter==len)
            System.out.println(counter-1);
        else if(counter<len)
            System.out.println(counter+1);
        else
            System.out.println(counter);
    }
}
                    

                        Solution in Python : 
                            
In  Python3 :







input()
a = [x for x in input().split()]
b = [x for x in input().split()]
aDict = dict()
bDict = dict()
for val in a:
    if val in aDict:
        aDict[val] += 1
    else:
        aDict[val] = 1
for val in b:
    if val in bDict:
        bDict[val] += 1
    else:
        bDict[val] = 1

total = 0
for val in aDict:
    while aDict[val] > 0 and val in bDict and bDict[val] > 0:
        total += 1
        aDict[val] -= 1
        bDict[val] -= 1

if total == len(a):
    print(total - 1)
else:
    print(total + 1)