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Merge pull request #199 from sir-gon/feature/frequency_queries
[Hacker Rank] Interview Preparation Kit: Dictionaries and Hashmaps: F...
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# [Dictionaries and Hashmaps: Frequency Queries](https://www.hackerrank.com/challenges/frequency-queries)
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- Difficulty: `#medium`
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- Category: `#ProblemSolvingIntermediate` `#dictionaries` `#hashmaps`
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You are given queries. Each query is of the form two integers described below:
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- `1 x`: Insert x in your data structure.
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- `2 y`: Delete one occurence of y from your data structure, if present.
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- `3 z`: Check if any integer is present whose frequency is exactly `z`.
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If yes, print `1` else `0`.
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The queries are given in the form of a 2-D array `queries` of
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size where `queries[i][0]` contains the operation,
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and `queries[i][0]` contains the data element.
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## Example
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The results of each operation are:
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```text
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Operation Array Output
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(1,1) [1]
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(2,2) [1]
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(3,2) 0
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(1,1) [1,1]
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(1,1) [1,1,1]
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(2,1) [1,1]
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(3,2) 1
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```
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Return an array with the output: [0, 1].
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## Function Description
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Complete the freqQuery function in the editor below.
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freqQuery has the following parameter(s):
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- `int queries[q][2]`: a 2-d array of integers
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## Returns
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- `int[]`: the results of queries of type `3`
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## Input Format
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The first line contains of an integer `q`, the number of queries.
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Each of the next `q` lines contains two space-separated integers,
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`queries[i][0]` and `queries[i][1]`.
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## Constraints
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- $ 1 \leq q \leq 10^5 $
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- $ 1 \leq x, y, z \leq 10^9 $
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- All $ queries[i][0] \isin \{1, 2, 3\} $
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- $ 1 \leq queries[i][1] \leq 10^9 $
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## Sample Input 0
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```text
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8
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1 5
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1 6
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3 2
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1 10
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1 10
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1 6
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2 5
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3 2
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```
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## Sample Output 0
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```text
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0
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1
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```
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## Explanation 0
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For the first query of type `3`, there is no integer
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whose frequency is `2` (`array = [5, 6]`).
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So answer is `0`.
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For the second query of type `3`, there are two integers
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in `array = [6, 10, 10, 6]` whose frequency is `2`(integer = `6` and `10`).
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So, the answer is `1`.
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## Sample Input 1
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```†ext
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4
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3 4
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2 1003
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1 16
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3 1
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```
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## Sample Output 1
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```†ext
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0
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1
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```
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## Explanation 1
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For the first query of type `3`, there is no integer of frequency `4`.
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The answer is `0`. For the second query of type `3`,
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there is one integer, `16` of frequency `1` so the answer is `1`.
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## Sample Input 2
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```text
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10
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1 3
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2 3
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3 2
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1 4
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1 5
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1 5
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1 4
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3 2
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2 4
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3 2
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```
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## Sample Output 2
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```text
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0
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1
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1
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```
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## Explanation 2
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When the first output query is run, the array is empty.
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We insert two `4`'s and two `5`'s before the second output query,
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`arr = [4, 5, 5, 4]` so there are two instances of elements occurring twice.
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We delete a `4` and run the same query.
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Now only the instances of `5` satisfy the query.
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// @link Problem definition [[docs/hackerrank/interview_preparation_kit/dictionaries_and_Dictionarys/frequency-queries.md]]
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namespace algorithm_exercises_csharp.hackerrank.interview_preparation_kit.dictionaries_and_hashmaps;
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using System.Diagnostics.CodeAnalysis;
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using System.Collections.Generic;
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public class FrequencyQueries
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{
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[ExcludeFromCodeCoverage]
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private FrequencyQueries() { }
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private static readonly long __INITIAL__ = 1L;
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private const int __INSERT__ = 1;
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private const int __DELETE__ = 2;
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private const int __SELECT__ = 3;
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private static readonly int __NOT_FOUND__ = 0;
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private static readonly int __FOUND__ = 1;
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readonly Dictionary<long, long> valueFreqs = [];
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readonly Dictionary<long, List<long>> freqDictionary = [];
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List<int> result = [];
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void insert(long value)
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{
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bool currentValueFreqCountExists = valueFreqs.TryGetValue(value, out long currentValueFreqCount);
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long newFreqCount;
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newFreqCount = !currentValueFreqCountExists ? __INITIAL__ : currentValueFreqCount + 1;
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valueFreqs[value] = newFreqCount;
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freqDictionary.TryGetValue(newFreqCount, out List<long>? newFreq);
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// delete current frequency
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if (currentValueFreqCountExists)
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{
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freqDictionary[currentValueFreqCount].Remove(value);
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if (freqDictionary[currentValueFreqCount].Count == 0)
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{
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freqDictionary.Remove(currentValueFreqCount);
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}
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}
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// add new frequency
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if (newFreq == null)
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{
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newFreq = [];
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newFreq.Add(value);
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freqDictionary[newFreqCount] = newFreq;
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}
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else
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{
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freqDictionary[newFreqCount].Add(value);
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}
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}
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void delete(long value)
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{
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bool currentValueFreqCountExists = valueFreqs.TryGetValue(value, out long currentValueFreqCount);
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long newFreqCount;
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newFreqCount = !currentValueFreqCountExists ? 0 : currentValueFreqCount - 1;
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if (newFreqCount > 0)
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{
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valueFreqs[value] = newFreqCount;
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freqDictionary.TryGetValue(newFreqCount, out List<long>? newFreq);
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// add new frequency
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if (newFreq == null)
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{
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newFreq = [];
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newFreq.Add(value);
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freqDictionary[newFreqCount] = newFreq;
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}
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else
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{
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freqDictionary[newFreqCount].Add(value);
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}
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}
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else
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{
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valueFreqs.Remove(value);
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}
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// delete current frequency
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if (currentValueFreqCountExists)
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{
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freqDictionary[currentValueFreqCount].Remove(value);
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if (freqDictionary[currentValueFreqCount].Count == 0)
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{
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freqDictionary.Remove(currentValueFreqCount);
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}
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}
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}
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void reset()
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{
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result = [];
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}
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void select(long value)
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{
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result.Add(freqDictionary.ContainsKey(value) ? __FOUND__ : __NOT_FOUND__);
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}
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static FrequencyQueries factory()
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{
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FrequencyQueries fq = new();
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fq.reset();
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return fq;
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}
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/**
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* FrequencyQueries.
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*/
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public static List<int> freqQuery(List<List<int>> queries)
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{
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FrequencyQueries fq = factory();
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foreach (List<int> query in queries)
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{
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int operation = query[0];
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long value = query[1];
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switch (operation)
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{
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case __INSERT__:
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fq.insert(value);
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break;
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case __DELETE__:
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fq.delete(value);
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break;
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case __SELECT__:
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fq.select(value);
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break;
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default:
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throw new InvalidOperationException($"Operation {operation} not supported");
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}
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}
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return fq.result;
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}
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}

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