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.

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)

{"mode":"full","isActive":false}

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;
    }
}

{"mode":"full","isActive":false}

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;
}

{"mode":"full","isActive":false}

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 %d\n",i,j,a[i],b[j]);
        if(a[i]==b[j])
        {
            //printf("j===%d\n",j);
            for(check=0,l=0;l<k;l++)
            {
               // printf("c[%d]==%d----j=%d\n",l,c[k],j);
                if(j==c[l])
                    check=1;
            }
            //printf("check---%d\n",check);
            if(check!=1)
            {
            c[k]=j;
           // printf("%d-vvv--\n",c[k]);
            k++;
           // printf("enter\n");
            count++;
            break;
            }
        }
    }
    if(count<n)
        count++;
    else if(count==n)
        count--;
    printf("%d",count);
    return 0;
}

{"mode":"full","isActive":false}