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abderahman
Asked: 2026-09-11T09:52:02-05:00 In: ,

How to compute Running Median in continuous data streams with O(log N) per tick?

In our telemetry service, financial tick prices arrive at ~10,000 events/second. We need an online algorithm that can output the exact running median at any moment.

Sorting the buffer on every tick is O(N log N), which is impossible at high frequency. What is the standard dual-heap architecture used to maintain running medians in real time?

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2 Answers

  1. Anonymous
    Begginer

    The classic, production-proven design for calculating running medians is the Dual-Heap Balancing Architecture (one Max-Heap and one Min-Heap).

    1. The Mental Model

    Imagine splitting all numbers you’ve seen so far into two equal halves:

    • The Lower Half (all numbers $le$ median): We store these in a Max-Heap. Why? Because we only care about the largest number in this half!
    • The Upper Half (all numbers $ge$ median): We store these in a Min-Heap. Why? Because we only care about the smallest number in this half!
       Lower Half (Max-Heap)          Upper Half (Min-Heap)
[1, 3, 5, 7, 9  -> (TOP: 9)]  <-->  [(TOP: 11) <- 11, 14, 18, 22]
                                     /
                        The Median is 
                     between 9 and 11!

    The median is ALWAYS right at the fingertips: either the top of the Max-Heap, or the average of the two tops!

    2. The Two Golden Invariants

    To make this work 100% reliably, you must maintain two invariants after every single number is added:

    1. Ordering Invariant: Every element in max_heap must be $le$ every element in min_heap. (If max_heap.top() > min_heap.top(), swap them).
    2. Size Balance Invariant: The size difference between the two heaps must never exceed 1: 0 <= len(max_heap) - len(min_heap) <= 1.

    Production Python 3.12 Implementation

    import heapq

class MedianFinder:
    def __init__(self):
        # Python heapq is a min-heap by default.
        # To simulate a max-heap, multiply values by -1.
        self.small = []  # Max-heap (stores inverted values)
        self.large = []  # Min-heap (stores normal values)

    def add_num(self, num: int) -> None:
        # Step 1: Push to max-heap (small half)
        heapq.heappush(self.small, -num)
        
        # Step 2: Ensure ordering invariant (all small <= all large)
        if self.small and self.large and (-self.small[0] > self.large[0]):
            val = -heapq.heappop(self.small)
            heapq.heappush(self.large, val)
            
        # Step 3: Ensure size invariant (len(small) can be at most 1 larger than len(large))
        if len(self.small) > len(self.large) + 1:
            val = -heapq.heappop(self.small)
            heapq.heappush(self.large, val)
        elif len(self.large) > len(self.small):
            val = heapq.heappop(self.large)
            heapq.heappush(self.small, -val)

    def find_median(self) -> float:
        if len(self.small) > len(self.large):
            return float(-self.small[0])
        return (-self.small[0] + self.large[0]) / 2.0

    Performance & Production Benchmarks

    • add_num() Time: O(log N). Pushing and popping from heaps of size N/2 takes ~15-20 CPU instructions.
    • find_median() Time: O(1). Simply peek at heap roots (index 0). Instantaneous!
    • Space Complexity: O(N) total memory to store the incoming stream numbers.
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  2. Abhishek
    Begginer

    Here is the clean C++20 Dual-Heap solution using std::priority_queue with std::greater<int> for the min-heap. This provides O(log N) insertion and O(1) median query.

    Modern C++20 Solution (Fully Runnable)

    #include <iostream>
#include <vector>
#include <queue>

class MedianFinder {
    // Max-heap stores the smaller half of numbers
    std::priority_queue<int> small;
    // Min-heap stores the larger half of numbers
    std::priority_queue<int, std::vector<int>, std::greater<int>> large;

public:
    void addNum(int num) {
        small.push(num);

        // Ensure small <= large
        if (!small.empty() && !large.empty() && small.top() > large.top()) {
            large.push(small.top());
            small.pop();
        }

        // Maintain size balance (small can have at most 1 more element)
        if (small.size() > large.size() + 1) {
            large.push(small.top());
            small.pop();
        } else if (large.size() > small.size()) {
            small.push(large.top());
            large.pop();
        }
    }

    double findMedian() const {
        if (small.size() > large.size()) {
            return static_cast<double>(small.top());
        }
        return (static_cast<double>(small.top()) + large.top()) / 2.0;
    }
};

int main() {
    MedianFinder mf;
    mf.addNum(5);
    mf.addNum(15);
    std::cout << "Median after [5, 15]: " << mf.findMedian() << "n"; // 10.0
    mf.addNum(1);
    std::cout << "Median after [5, 15, 1]: " << mf.findMedian() << "n"; // 5.0
    mf.addNum(3);
    std::cout << "Median after [5, 15, 1, 3]: " << mf.findMedian() << "n"; // 4.0
    return 0;
}

    Complexity: O(log N) per tick and O(1) for median queries.

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