**Beautiful Pairs**

### Problem Statement :

You are given two arrays, and , both containing integers. A pair of indices is beautiful if the element of array is equal to the element of array . In other words, pair is beautiful if and only if . A set containing beautiful pairs is called a beautiful set. A beautiful set is called pairwise disjoint if for every pair belonging to the set there is no repetition of either or values. For instance, if and the beautiful set is not pairwise disjoint as there is a repetition of , that is . Your task is to change exactly element in so that the size of the pairwise disjoint beautiful set is maximum You are given two arrays, and , both containing integers. A pair of indices is beautiful if the element of array is equal to the element of array . In other words, pair is beautiful if and only if . A set containing beautiful pairs is called a beautiful set. Output Format Determine and print the maximum possible number of pairwise disjoint beautiful pairs. Note: You must first change element in , and your choice of element must be optimal. A beautiful set is called pairwise disjoint if for every pair belonging to the set there is no repetition of either or values. For instance, if and the beautiful set is not pairwise disjoint as there is a repetition of , that is . Your task is to change exactly element in so that the size of the pairwise disjoint beautiful set is maximum

### Solution :

` ````
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)
```

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