[Hacker Rank] Interview Preparation Kit: Dictionaries and Hashmaps: F…

docs/hackerrank/interview_preparation_kit/dictionaries_and_hashmaps/frequency-queries.md

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+# [Dictionaries and Hashmaps: Frequency Queries](https://www.hackerrank.com/challenges/frequency-queries)
+
+- Difficulty:  `#medium`
+- Category: `#ProblemSolvingIntermediate` `#dictionaries` `#hashmaps`
+
+You are given  queries. Each query is of the form two integers described below:
+
+- `1 x`: Insert x in your data structure.
+- `2 y`: Delete one occurence of y from your data structure, if present.
+- `3 z`: Check if any integer is present whose frequency is exactly `z`.
+If yes, print `1` else `0`.
+
+The queries are given in the form of a 2-D array `queries` of
+size where `queries[i][0]` contains the operation,
+and `queries[i][0]` contains the data element.
+
+## Example
+
+The results of each operation are:
+
+```text
+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: [0, 1].
+
+## Function Description
+
+Complete the freqQuery function in the editor below.
+
+freqQuery has the following parameter(s):
+
+- `int queries[q][2]`: a 2-d array of integers
+
+## Returns
+
+- `int[]`: the results of queries of type `3`
+
+## Input Format
+
+The first line contains of an integer `q`, the number of queries.
+
+Each of the next `q` lines contains two space-separated integers,
+`queries[i][0]` and `queries[i][1]`.
+
+## Constraints
+
+- $ 1 \leq q \leq 10^5 $
+- $ 1 \leq x, y, z \leq 10^9 $
+- All $ queries[i][0] \isin \{1, 2, 3\} $
+- $ 1 \leq queries[i][1] \leq 10^9 $
+
+## Sample Input 0
+
+```text
+8
+1 5
+1 6
+3 2
+1 10
+1 10
+1 6
+2 5
+3 2
+```
+
+## Sample Output 0
+
+```text
+0
+1
+```
+
+## Explanation 0
+
+For the first query of type `3`, there is no integer
+whose frequency is `2` (`array = [5, 6]`).
+So answer is `0`.
+
+For the second query of type `3`, there are two integers
+in `array = [6, 10, 10, 6]` whose frequency is `2`(integer = `6` and `10`).
+So, the answer is `1`.
+
+## Sample Input 1
+
+```†ext
+4
+3 4
+2 1003
+1 16
+3 1
+```
+
+## Sample Output 1
+
+```†ext
+0
+1
+```
+
+## Explanation 1
+
+For the first query of type `3`, there is no integer of frequency `4`.
+The answer is `0`. For the second query of type `3`,
+there is one integer, `16` of frequency `1` so the answer is `1`.
+
+## Sample Input 2
+
+```text
+10
+1 3
+2 3
+3 2
+1 4
+1 5
+1 5
+1 4
+3 2
+2 4
+3 2
+```
+
+## Sample Output 2
+
+```text
+0
+1
+1
+
+```
+
+## Explanation 2
+
+When the first output query is run, the array is empty.
+We insert two `4`'s and two `5`'s before the second output query,
+`arr = [4, 5, 5, 4]` so there are two instances of elements occurring twice.
+We delete a `4` and run the same query.
+Now only the instances of `5` satisfy the query.

src/algorithm_exercises_csharp/hackerrank/interview_preparation_kit/dictionaries_and_hashmaps/FrequencyQueries.cs

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+// @link Problem definition [[docs/hackerrank/interview_preparation_kit/dictionaries_and_Dictionarys/frequency-queries.md]]
+
+namespace algorithm_exercises_csharp.hackerrank.interview_preparation_kit.dictionaries_and_hashmaps;
+
+using System.Diagnostics.CodeAnalysis;
+using System.Collections.Generic;
+
+public class FrequencyQueries
+{
+  [ExcludeFromCodeCoverage]
+  private FrequencyQueries() { }
+
+  private static readonly long __INITIAL__ = 1L;
+
+  private const int __INSERT__ = 1;
+  private const int __DELETE__ = 2;
+  private const int __SELECT__ = 3;
+
+  private static readonly int __NOT_FOUND__ = 0;
+  private static readonly int __FOUND__ = 1;
+
+  readonly Dictionary<long, long> valueFreqs = [];
+  readonly Dictionary<long, List<long>> freqDictionary = [];
+  List<int> result = [];
+
+  void insert(long value)
+  {
+    bool currentValueFreqCountExists = valueFreqs.TryGetValue(value, out long currentValueFreqCount);
+
+    long newFreqCount;
+
+    newFreqCount = !currentValueFreqCountExists ? __INITIAL__ : currentValueFreqCount + 1;
+    valueFreqs[value] = newFreqCount;
+
+    freqDictionary.TryGetValue(newFreqCount, out List<long>? newFreq);
+
+    // delete current frequency
+    if (currentValueFreqCountExists)
+    {
+      freqDictionary[currentValueFreqCount].Remove(value);
+      if (freqDictionary[currentValueFreqCount].Count == 0)
+      {
+        freqDictionary.Remove(currentValueFreqCount);
+      }
+    }
+
+    // add new frequency
+    if (newFreq == null)
+    {
+      newFreq = [];
+      newFreq.Add(value);
+      freqDictionary[newFreqCount] = newFreq;
+    }
+    else
+    {
+      freqDictionary[newFreqCount].Add(value);
+    }
+  }
+
+  void delete(long value)
+  {
+    bool currentValueFreqCountExists = valueFreqs.TryGetValue(value, out long currentValueFreqCount);
+
+    long newFreqCount;
+
+    newFreqCount = !currentValueFreqCountExists ? 0 : currentValueFreqCount - 1;
+    if (newFreqCount > 0)
+    {
+      valueFreqs[value] = newFreqCount;
+
+      freqDictionary.TryGetValue(newFreqCount, out List<long>? newFreq);
+
+      // add new frequency
+      if (newFreq == null)
+      {
+        newFreq = [];
+        newFreq.Add(value);
+        freqDictionary[newFreqCount] = newFreq;
+      }
+      else
+      {
+        freqDictionary[newFreqCount].Add(value);
+      }
+    }
+    else
+    {
+      valueFreqs.Remove(value);
+    }
+
+    // delete current frequency
+    if (currentValueFreqCountExists)
+    {
+      freqDictionary[currentValueFreqCount].Remove(value);
+      if (freqDictionary[currentValueFreqCount].Count == 0)
+      {
+        freqDictionary.Remove(currentValueFreqCount);
+      }
+    }
+  }
+
+  void reset()
+  {
+    result = [];
+  }
+
+  void select(long value)
+  {
+    result.Add(freqDictionary.ContainsKey(value) ? __FOUND__ : __NOT_FOUND__);
+  }
+
+  static FrequencyQueries factory()
+  {
+    FrequencyQueries fq = new();
+    fq.reset();
+
+    return fq;
+  }
+
+  /**
+   * FrequencyQueries.
+   */
+  public static List<int> freqQuery(List<List<int>> queries)
+  {
+    FrequencyQueries fq = factory();
+
+    foreach (List<int> query in queries)
+    {
+      int operation = query[0];
+      long value = query[1];
+
+      switch (operation)
+      {
+        case __INSERT__:
+          fq.insert(value);
+
+          break;
+        case __DELETE__:
+          fq.delete(value);
+
+          break;
+        case __SELECT__:
+          fq.select(value);
+          break;
+        default:
+          throw new InvalidOperationException($"Operation {operation} not supported");
+      }
+    }
+
+    return fq.result;
+  }
+}