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LeetCode 3369: Design an Array Statistics Tracker Solution

Master LeetCode problem 3369 (Design an Array Statistics Tracker ), a hard 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.

Problem Explanation

Explanation:

To design an Array Statistics Tracker, we can use a combination of data structures to efficiently keep track of various statistics of an array. We can use a hashmap to store the frequency of each element in the array and a min-heap and max-heap to maintain the median of the array.

Here are the steps:

  1. Create a hashmap to store the frequency of each element in the array.
  2. Create a max-heap to store the smaller half of the array elements.
  3. Create a min-heap to store the larger half of the array elements.
  4. Maintain the heaps such that the size difference between the two heaps is at most 1, to efficiently calculate the median.
  5. Implement methods to add elements to the array, remove elements from the array, and calculate the median of the array.

Time complexity:

  • Adding an element: O(log n) for updating heaps and hashmap.
  • Removing an element: O(log n) for updating heaps and hashmap.
  • Calculating median: O(1).

Space complexity: O(n) for storing the elements in the heaps and hashmap.

:

Solution Code

import java.util.*;

class ArrayStatisticsTracker {
    Map<Integer, Integer> freqMap;
    PriorityQueue<Integer> maxHeap;
    PriorityQueue<Integer> minHeap;

    public ArrayStatisticsTracker() {
        freqMap = new HashMap<>();
        maxHeap = new PriorityQueue<>(Collections.reverseOrder());
        minHeap = new PriorityQueue<>();
    }

    public void addNum(int num) {
        freqMap.put(num, freqMap.getOrDefault(num, 0) + 1);
        maxHeap.offer(num);
        minHeap.offer(maxHeap.poll());
        
        if (maxHeap.size() < minHeap.size()) {
            maxHeap.offer(minHeap.poll());
        }
    }

    public void removeNum(int num) {
        if (!freqMap.containsKey(num)) {
            return;
        }
        
        freqMap.put(num, freqMap.get(num) - 1);
        while (!maxHeap.isEmpty() && freqMap.getOrDefault(maxHeap.peek(), 0) == 0) {
            maxHeap.poll();
        }
        while (!minHeap.isEmpty() && freqMap.getOrDefault(minHeap.peek(), 0) == 0) {
            minHeap.poll();
        }
    }

    public double findMedian() {
        if (maxHeap.size() == minHeap.size()) {
            return (double) (maxHeap.peek() + minHeap.peek()) / 2;
        } else {
            return (double) maxHeap.peek();
        }
    }
}

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