[Hackerrank] missing numbers

Numeros, the artist, had two listsAAndB, Such that,BWas a permutationA. Numeros was very proud of these lists. Unfortunately, while transporting them from one exhibition to another, some numbers from ListAGot left out. Can you find out the numbers missing fromA?

Notes

• If a number occurs multiple times in the lists, you must ensure that the frequency of that number in both the lists is the same. if that is not the case, then it is also a missing number.
• You have to print all the missing numbers in ascending order.
• Print each missing number once, even if it is missing multiple times.
• The difference between maximum and minimum number in the listBIs less than or equal to 100.

Input Format 
There will be four lines of input:

N-The size of the first list
This is followedNSpace separated integers that make up the first list.
M-The size of the second list
This is followedMSpace separated integers that make up the second list.

Output Format 
Output the missing numbers in ascending order

Constraints 
1 <=N, m<= 1000010
1 <=X<= 10000,Xε B 
Xmax-Xmin<101

 

Question: set two arrays because the range of X is 1 ~ Between 10000, you only need to open two 10001 arrays to record the number of elements in A and B respectively, and then compare the two arrays.

The Code is as follows:

 1 import java.util.*; 2  3 public class Solution { 4     public static void main(String[] args) { 5         Scanner in = new Scanner(System.in); 6         int[] CountA = new int[10005]; 7         int[] CountB = new int[10005]; 8          9         int n = in.nextInt();10         int[] a = new int[n];11         for(int i = 0;i < n;i++){12             a[i]=in.nextInt();13             CountA[a[i]]++;14         }15         16         int m = in.nextInt();17         int[] b = new int[m];18         for(int i = 0;i < m;i++){19             b[i]=in.nextInt();20             CountB[b[i]]++;21         }22         23         for(int i = 1;i <= 10000;i++){24             if(CountB[i]>CountA[i] )25                 System.out.printf("%d ", i);26         }27         System.out.println();28         29         30         31     }32 }