题目

You are given queries. Each query is of the form two integers described below:

• : Insert x in your data structure.
• : Delete one occurence of y from your data structure, if present.
• : Check if any integer is present whose frequency is exactly . If yes, print 1 else 0.

The queries are given in the form of a 2-D array of size where contains the operation, and contains the data element. For example, you are given array . The results of each operation are:

Operation Array Output
(1,1) [1]
(2,2) [1]
(3,2) 0
(1,1) [1,1]
(1,1) [1,1,1]
(2,1) [1,1]
(3,2) 1
Return an array with the output: .

Function Description

Complete the freqQuery function in the editor below. It must return an array of integers where each element is a if there is at least one element value with the queried number of occurrences in the current array, or 0 if there is not.

freqQuery has the following parameter(s):

queries: a 2-d array of integers
Input Format

The first line contains of an integer , the number of queries.
Each of the next lines contains two integers denoting the 2-d array .

Constraints

All
Output Format

Return an integer array consisting of all the outputs of queries of type .

Sample Input 0

8
1 5
1 6
3 2
1 10
1 10
1 6
2 5
3 2
Sample Output 0

0
1
Explanation 0

For the first query of type , there is no integer whose frequency is (). So answer is .
For the second query of type , there are two integers in whose frequency is (integers = and ). So, the answer is .

Sample Input 1

4
3 4
2 1003
1 16
3 1
Sample Output 1

0
1
Explanation 1

For the first query of type , there is no integer of frequency . The answer is .
For the second query of type , there is one integer, of frequency so the answer is .

Sample Input 2

10
1 3
2 3
3 2
1 4
1 5
1 5
1 4
3 2
2 4
3 2
Sample Output 2

0
1
1
Explanation 2

When the first output query is run, the array is empty. We insert two 's and two 's before the second output query, so there are two instances of elements occurring twice. We delete a and run the same query. Now only the instances of satisfy the query.

题目解析

相对比较简单的一个题目，利用字典根据不同的action处理即可，逻辑绕在出现次数的加减上面，比如出现5次减一那就意味着出现4次的加一

Answer

#!/bin/python3

import math
import os
import random
import re
import sys

# Complete the freqQuery function below.
def freqQuery(queries):
    count_dic = {}
    freq_dic = {}
    for case, data in queries:
        if case == 1:
            count = count_dic.setdefault(data, 0)
            freq = freq_dic.setdefault(count, 1)
            count_dic[data] += 1
            freq_dic.setdefault(count+1, 0)
            freq_dic[count+1] += 1
            freq_dic[count] -= 1
        elif case == 2:
            count = count_dic.setdefault(data, 0)
            freq = freq_dic.setdefault(count, 1)
            if count > 0:
                count_dic[data] -= 1
                freq_dic.setdefault(count-1, 0)
                freq_dic[count-1] += 1
                freq_dic[count] -= 1
        else:
            yield 1 if freq_dic.get(data, 0) > 0 else 0





if __name__ == '__main__':
    fptr = open(os.environ['OUTPUT_PATH'], 'w')

    q = int(input().strip())

    queries = []

    for _ in range(q):
        queries.append(list(map(int, input().rstrip().split())))

    ans = freqQuery(queries)

    fptr.write('\n'.join(map(str, ans)))
    fptr.write('\n')

    fptr.close()