2671. Frequency Tracker

Problem

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 occurrence 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 add, deleteOne, and hasFrequency in total.

Solution (Java)

class FrequencyTracker {
    int[] b;
    HashMap<Integer,Integer> map;
    public FrequencyTracker() {
        b= new int[200001];
        map = new HashMap<>();
    }

    public void add(int number) {
        int x = map.getOrDefault(number,0);
        if(x!=0){
            b[x]--;

        }
        b[x+1]++;
        map.put(number,x+1);

    }

    public void deleteOne(int number) {
        int x =map.getOrDefault(number,0);
        if(x==0){ // don't do anything

        }
        else {
            b[x]--;
            b[x-1]++;
            map.put(number,x-1);
        }
    }

    public boolean hasFrequency(int frequency) {
        return b[frequency]>0;
    }
}

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

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).