Find Median from Data Stream

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() initializes 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 1:

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 (i.e., (1 + 2) / 2) medianFinder.addNum(3); // arr[1, 2, 3] medianFinder.findMedian(); // return 2.0

Constraints:

• -105 <= num <= 105
• There will be at least one element in the data structure before calling findMedian.
• At most 5 * 104 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 optimize your solution?
• If 99% of all integer numbers from the stream are in the range [0, 100], how would you optimize your solution?

Approach: Two Heaps (Max-Heap + Min-Heap)

Algorithm

1. Keep two heaps: a max-heap for the smaller half of the numbers and a min-heap for the larger half
2. addNum: push into the max-heap, then move the largest element to the min-heap; rebalance so the two heaps differ in size by at most one
3. findMedian: if the sizes are equal, average the two roots; otherwise return the root of the larger heap

Time & Space Complexity

• Time Complexity: O(log n) per addNum; O(1) per findMedian
• Space Complexity: O(n) - all stream elements are stored

Java Implementation

import java.util.Collections;
import java.util.PriorityQueue;

public class MedianFinder {

    private final PriorityQueue<Integer> maxHeap;
    private final PriorityQueue<Integer> minHeap;

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

    public void addNum(int num) {
        maxHeap.offer(num);
        minHeap.offer(maxHeap.poll());

        if (maxHeap.size() < minHeap.size()) {
            maxHeap.offer(minHeap.poll());
        }
    }

    public double findMedian() {
        if (maxHeap.size() > minHeap.size()) {
            return maxHeap.peek();
        }
        return (maxHeap.peek() + minHeap.peek()) / 2.0;
    }

    // Test method
    public static void main(String[] args) {
        MedianFinder medianFinder = new MedianFinder();
        medianFinder.addNum(1);
        medianFinder.addNum(2);
        System.out.println(medianFinder.findMedian()); // return 1.5
        medianFinder.addNum(3);
        System.out.println(medianFinder.findMedian()); // return 2.0
    }
}

Example Walkthrough

Inserting 1, 2, 3:

1. addNum(1): maxHeap={1}, sizes equal
2. addNum(2): push 2 to maxHeap -> maxHeap={2,1}; move max to minHeap -> maxHeap={1}, minHeap={2}. Even sizes, median=(1+2)/2=1.5
3. addNum(3): push to maxHeap -> maxHeap={3,1}; move max to minHeap -> maxHeap={1}, minHeap={2,3}; rebalance since maxHeap < minHeap -> maxHeap={2,1}, minHeap={3}. maxHeap larger, median=2.0

Key Points

1. Invariant: maxHeap holds the smaller half, minHeap holds the larger half, and sizes differ by at most one
2. Top of Heaps: The median candidates are always the roots of the two heaps
3. Balancing Step: The maxHeap is never smaller than minHeap, keeping the median in maxHeap for odd sizes
4. Follow-Up: For bounded ranges [0,100], a frequency array with prefix sums gives O(100) operations
5. Precision: Averaging in double arithmetic keeps results within the 10-5 tolerance