LeetCode 2671: Frequency Tracker Solution

Master LeetCode problem 2671 (Frequency Tracker), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2671. Frequency Tracker

Medium

Hash Table Design

Problem Explanation

Explanation

To solve this problem, we can use a HashMap to keep track of the frequency of each number in the data structure. We will also maintain a separate HashMap to keep track of the frequency of each frequency value. When adding a number, we update both HashMaps accordingly. When deleting one occurrence of a number, we update the HashMaps. To check if a particular frequency exists, we simply query the second HashMap.

Time Complexity

Adding a number: O(1)
Deleting one occurrence of a number: O(1)
Checking if a frequency exists: O(1)

Space Complexity

O(n), where n is the number of unique numbers in the data structure.

Solution Code

import java.util.HashMap;

class FrequencyTracker {
    private HashMap<Integer, Integer> numToFreq;
    private HashMap<Integer, Integer> freqToCount;

    public FrequencyTracker() {
        numToFreq = new HashMap<>();
        freqToCount = new HashMap<>();
    }

    public void add(int number) {
        int prevFreq = numToFreq.getOrDefault(number, 0);
        numToFreq.put(number, prevFreq + 1);
        
        if (prevFreq > 0) {
            int prevCount = freqToCount.get(prevFreq);
            if (prevCount == 1) {
                freqToCount.remove(prevFreq);
            } else {
                freqToCount.put(prevFreq, prevCount - 1);
            }
        }
        
        int newFreq = prevFreq + 1;
        freqToCount.put(newFreq, freqToCount.getOrDefault(newFreq, 0) + 1);
    }

    public void deleteOne(int number) {
        if (!numToFreq.containsKey(number)) return;
        
        int prevFreq = numToFreq.get(number);
        if (prevFreq == 1) {
            numToFreq.remove(number);
        } else {
            numToFreq.put(number, prevFreq - 1);
            int prevCount = freqToCount.get(prevFreq);
            if (prevCount == 1) {
                freqToCount.remove(prevFreq);
            } else {
                freqToCount.put(prevFreq, prevCount - 1);
            }
            freqToCount.put(prevFreq - 1, freqToCount.getOrDefault(prevFreq - 1, 0) + 1);
        }
    }

    public boolean hasFrequency(int frequency) {
        return freqToCount.containsKey(frequency);
    }
}

Try It Yourself

Loading code editor...

Frequently Asked Questions

How to solve LeetCode 2671 (Frequency Tracker)?

This page provides optimized solutions for LeetCode problem 2671 (Frequency Tracker) in Java, C++, and Python, along with a detailed explanation and an interactive code editor to test your code.

What is the time complexity of LeetCode 2671 (Frequency Tracker)?

The time complexity for LeetCode 2671 (Frequency Tracker) varies by solution. Check the detailed explanation section for specific complexities in Java, C++, and Python implementations.

Can I run code for LeetCode 2671 on DevExCode?

Yes, DevExCode provides an interactive code editor where you can write, test, and run your code for LeetCode 2671 in Java, C++, or Python.

Back to LeetCode Solutions