Skip to main content

Frequncy Tracker Leetcode

 2671. Frequency Tracker

Design a data structure that keeps track of the values in it and answers some queries regarding their frequencies.

Implement the FrequencyTracker class.

  • FrequencyTracker(): Initializes the FrequencyTracker object with an empty array initially.
  • void add(int number): Adds number to the data structure.
  • void deleteOne(int number): Deletes one occurence of number from the data structure. The data structure may not contain number, and in this case nothing is deleted.
  • bool hasFrequency(int frequency): Returns true if there is a number in the data structure that occurs frequency number of times, otherwise, it returns false.

 

Example 1:

Input
["FrequencyTracker", "add", "add", "hasFrequency"]
[[], [3], [3], [2]]
Output
[null, null, null, true]

Explanation
FrequencyTracker frequencyTracker = new FrequencyTracker();
frequencyTracker.add(3); // The data structure now contains [3]
frequencyTracker.add(3); // The data structure now contains [3, 3]
frequencyTracker.hasFrequency(2); // Returns true, because 3 occurs twice

Example 2:

Input
["FrequencyTracker", "add", "deleteOne", "hasFrequency"]
[[], [1], [1], [1]]
Output
[null, null, null, false]

Explanation
FrequencyTracker frequencyTracker = new FrequencyTracker();
frequencyTracker.add(1); // The data structure now contains [1]
frequencyTracker.deleteOne(1); // The data structure becomes empty []
frequencyTracker.hasFrequency(1); // Returns false, because the data structure is empty

Example 3:

Input
["FrequencyTracker", "hasFrequency", "add", "hasFrequency"]
[[], [2], [3], [1]]
Output
[null, false, null, true]

Explanation
FrequencyTracker frequencyTracker = new FrequencyTracker();
frequencyTracker.hasFrequency(2); // Returns false, because the data structure is empty
frequencyTracker.add(3); // The data structure now contains [3]
frequencyTracker.hasFrequency(1); // Returns true, because 3 occurs once

 

Constraints:

  • 1 <= number <= 105
  • 1 <= frequency <= 105
  • At most, 2 * 105 calls will be made to adddeleteOne, and hasFrequency in total.

Solution : 

class FrequencyTracker {
public:
    unordered_map<int, int>map, frequency ;
    FrequencyTracker() {
        // FrequencyTracker frequencyTracker = new FrequencyTracker();
    }
 
    void add(int number) {
       
        map[number]++;
        int x = map[number];
        frequency[x-1] --;
        frequency[x] ++ ;
    }
 
    void deleteOne(int number) {
        if(map[number])
        {map[number]--;
            int x = map[number];
          frequency[x] ++ ;;
        frequency[x+1] --;
        }
    }
 
    bool hasFrequency(int k) {
     
        if(frequency[k])
            return true;
        return false;
         
    }
};

/**
* Your FrequencyTracker object will be instantiated and called as such:
* FrequencyTracker* obj = new FrequencyTracker();
* obj->add(number);
* obj->deleteOne(number);
* bool param_3 = obj->hasFrequency(frequency);
*/


Comments

Popular posts from this blog

leetcode 48 solution

  48 .  Rotate Image You are given an  n x n  2D  matrix  representing an image, rotate the image by  90  degrees (clockwise). You have to rotate the image  in-place , which means you have to modify the input 2D matrix directly.  DO NOT  allocate another 2D matrix and do the rotation.   Example 1: Input: matrix = [[1,2,3],[4,5,6],[7,8,9]] Output: [[7,4,1],[8,5,2],[9,6,3]] Example 2: Input: matrix = [[5,1,9,11],[2,4,8,10],[13,3,6,7],[15,14,12,16]] Output: [[15,13,2,5],[14,3,4,1],[12,6,8,9],[16,7,10,11]]   Constraints: n == matrix.length == matrix[i].length 1 <= n <= 20 -1000 <= matrix[i][j] <= 1000 solution: class Solution { public:     void swap(int& a , int &b)     {         int c ;         c = a;         a = b;         b = c;     }     void transpose (vector<vector<int>...

Regular Expression Matching Leetcode Solution

Regular Expression Matching Given an input string s and a pattern p, implement regular expression matching with support for '.' and '*' where: '.' Matches any single character.​​​​ '*' Matches zero or more of the preceding element. The matching should cover the entire input string (not partial). Example 1: Input: s = "aa", p = "a"  Output: false  Explanation: "a" does not match the entire string "aa". Example 2: Input: s = "aa", p = "a*"  Output: true  Explanation: '*' means zero or more of the preceding element, 'a'. Therefore, by repeating 'a' once, it becomes "aa". Example 3: Input: s = "ab", p = ".*"  Output: true  Explanation: ".*" means "zero or more (*) of any character (.)". Constraints: 1 <= s.length <= 20 1 <= p.length <= 20 s contains only lowercase English letters. p contains only lowercase Englis...

2485. Find the Pivot Integer | Binary search

  Given a positive integer   n , find the   pivot integer   x   such that: The sum of all elements between  1  and  x  inclusively equals the sum of all elements between  x  and  n  inclusively. Return  the pivot integer  x . If no such integer exists, return  -1 . It is guaranteed that there will be at most one pivot index for the given input.   Example 1: Input: n = 8 Output: 6 Explanation: 6 is the pivot integer since: 1 + 2 + 3 + 4 + 5 + 6 = 6 + 7 + 8 = 21. Example 2: Input: n = 1 Output: 1 Explanation: 1 is the pivot integer since: 1 = 1. Example 3: Input: n = 4 Output: -1 Explanation: It can be proved that no such integer exist.   Constraints: 1 <= n <= 1000 Solution : class Solution { publ ic:     int pivotInteger( int n ) {         int sum = (( n )*( n + 1 ))/ 2 ;         int i = 1 ;         int j =...