Find Median from Data Stream

LeetCode #295 — Hard

The median is the middle value in an ordered integer list. If the size of the list is even, there is no middle value, and the median is the mean of the two middle values.

For example, for arr = [2, 3, 4], the median is 3. For example, for arr = [2, 3], the median is (2 + 3) / 2 = 2.5.

Implement the MedianFinder class:

• MedianFinder() initialises the MedianFinder object.
• void addNum(int num) adds the integer num from the data stream to the data structure.
• double findMedian() returns the median of all elements so far. Answers within 10^-5 of the actual answer will be accepted.

Example

Input:
    ["MedianFinder", "addNum", "addNum", "findMedian", "addNum", "findMedian"]
    [[],             [1],      [2],      [],           [3],      []]

Output:
    [null, null, null, 1.5, null, 2.0]

Explanation:
    MedianFinder medianFinder = new MedianFinder();
    medianFinder.addNum(1);    // arr = [1]
    medianFinder.addNum(2);    // arr = [1, 2]
    medianFinder.findMedian(); // return 1.5
    medianFinder.addNum(3);    // arr = [1, 2, 3]
    medianFinder.findMedian(); // return 2.0

Constraints

• -10^5 <= num <= 10^5
• There will be at least one element in the data structure before calling findMedian.
• At most 5 * 10^4 calls will be made to addNum and findMedian.

Follow-up

• If all integer numbers from the stream are in the range [0, 100], how would you optimise it?
• If 99% of all integer numbers from the stream are in the range [0, 100], how would you optimise it?

What to practise

• Two heaps: a max-heap for the lower half, a min-heap for the upper half.
• Invariant: len(low) == len(high) or len(low) == len(high) + 1.
• On addNum: push to low, then move low.top() to high; if len(high) > len(low), move high.top() back to low.
• On findMedian: if sizes equal → mean of tops; else → top of low.
• Python: simulate max-heap by negating values pushed into heapq.