HackerRank Beautiful Pairs problem solution

In this HackerRank Beautiful Pairs problem solution, we have given two arrays A and B, and both containing N integers and we need to change exactly one element in B so that the size of the pairwise disjoint is beautiful set is maximum.

Function Description

Complete the beautifulPairs function in the editor below. It should return an integer that represents the maximum number of pairwise disjoint beautiful pairs that can be formed.

beautifulPairs has the following parameters:

• A: an array of integers
• B: an array of integers

HackerRank Beautiful Pairs problem solution in Python.

n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))

res = 0
for i in range(n):
    for j in range (n):
        if a[i] == b[j] and a[i] > 0:
            res += 1
            a[i] = 0 #just set to 0 to be not used anymore
            b[j] = 0

if res < n:
    res += 1
else:
    res -= 1

print (res)

Beautiful Pairs problem solution in 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[] A = readArray(in, N);
        int[] B = readArray(in, N);
        Arrays.sort(A);
        Arrays.sort(B);
        int i = 0;
        int j = 0;
        int count = 0;
        while (i < N && j < N) {
            if (A[i] < B[j]) i++;
            else if (A[i] > B[j]) j++;
            else {
                i++;
                j++;
                count++;
            }
        }
        if (count < N) count++;
        else count--;
        System.out.println(count);
    }
    
    public static int[] readArray(Scanner in, int N) {
        int[] ar = new int[N];
        for (int i = 0; i < N; i++) {
            ar[i] = in.nextInt();
        }
        return ar;
    }
}

Problem solution in C++.

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;


int main() {
    int n;
    cin>>n;
    vector<int> a(1000, 0),b(1000, 0);
    for(int i = 0;i < n;i++)
        {
        int c;
        cin>>c;
        a[c]++;
    }
    for(int i = 0;i < n;i++)
        {
        int c;
        cin>>c;
        b[c]++;
    }
    int r = 0;
    for(int i = 1;i <= 1000;i++)
        {
        r += min(a[i], b[i]);
    }
    cout<<(r==n?n-1:r+1);
    return 0;
}

Problem solution in C.

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

int main() {

    int n,i,j,l,k,count=0,check=0;
    scanf("%d",&n);
    int a[n],b[n];
    int c[n];
    
    //t ad[n],bd[n];
    for(i=0;i<n;i++)
        scanf("%d",&a[i]);
    for(i=0;i<n;i++)
        scanf("%d",&b[i]);
    for(k=0,i=0;i<n;i++)
    for(j=0;j<n;j++){
        //printf("i=%d j=%d   %d %dn",i,j,a[i],b[j]);
        if(a[i]==b[j])
        {
            //printf("j===%dn",j);
            for(check=0,l=0;l<k;l++)
            {
               // printf("c[%d]==%d----j=%dn",l,c[k],j);
                if(j==c[l])
                    check=1;
            }
            //printf("check---%dn",check);
            if(check!=1)
            {
            c[k]=j;
           // printf("%d-vvv--n",c[k]);
            k++;
           // printf("entern");
            count++;
            break;
            }
        }
    }
    if(count<n)
        count++;
    else if(count==n)
        count--;
    printf("%d",count);
    return 0;
}