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295. Find Median from Data Stream

Description

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?

Solutions

Solution 1: Min Heap and Max Heap (Priority Queue)

We can use two heaps to maintain all the elements, a min heap $\textit{minQ}$ and a max heap $\textit{maxQ}$, where the min heap $\textit{minQ}$ stores the larger half, and the max heap $\textit{maxQ}$ stores the smaller half.

When calling the addNum method, we first add the element to the max heap $\textit{maxQ}$, then pop the top element of $\textit{maxQ}$ and add it to the min heap $\textit{minQ}$. If at this time the size difference between $\textit{minQ}$ and $\textit{maxQ}$ is greater than $1$, we pop the top element of $\textit{minQ}$ and add it to $\textit{maxQ}$. The time complexity is $O(\log n)$.

When calling the findMedian method, if the size of $\textit{minQ}$ is equal to the size of $\textit{maxQ}$, it means the total number of elements is even, and we can return the average value of the top elements of $\textit{minQ}$ and $\textit{maxQ}$; otherwise, we return the top element of $\textit{minQ}$. The time complexity is $O(1)$.

The space complexity is $O(n)$, where $n$ is the number of elements.

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class MedianFinder:

    def __init__(self):
        self.minq = []
        self.maxq = []

    def addNum(self, num: int) -> None:
        heappush(self.minq, -heappushpop(self.maxq, -num))
        if len(self.minq) - len(self.maxq) > 1:
            heappush(self.maxq, -heappop(self.minq))

    def findMedian(self) -> float:
        if len(self.minq) == len(self.maxq):
            return (self.minq[0] - self.maxq[0]) / 2
        return self.minq[0]


# Your MedianFinder object will be instantiated and called as such:
# obj = MedianFinder()
# obj.addNum(num)
# param_2 = obj.findMedian()

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class MedianFinder {
    private PriorityQueue<Integer> minQ = new PriorityQueue<>();
    private PriorityQueue<Integer> maxQ = new PriorityQueue<>(Collections.reverseOrder());

    public MedianFinder() {
    }

    public void addNum(int num) {
        maxQ.offer(num);
        minQ.offer(maxQ.poll());
        if (minQ.size() - maxQ.size() > 1) {
            maxQ.offer(minQ.poll());
        }
    }

    public double findMedian() {
        return minQ.size() == maxQ.size() ? (minQ.peek() + maxQ.peek()) / 2.0 : minQ.peek();
    }
}

/**
 * Your MedianFinder object will be instantiated and called as such:
 * MedianFinder obj = new MedianFinder();
 * obj.addNum(num);
 * double param_2 = obj.findMedian();
 */

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class MedianFinder {
public:
    MedianFinder() {
    }

    void addNum(int num) {
        maxQ.push(num);
        minQ.push(maxQ.top());
        maxQ.pop();

        if (minQ.size() > maxQ.size() + 1) {
            maxQ.push(minQ.top());
            minQ.pop();
        }
    }

    double findMedian() {
        return minQ.size() == maxQ.size() ? (minQ.top() + maxQ.top()) / 2.0 : minQ.top();
    }

private:
    priority_queue<int> maxQ;
    priority_queue<int, vector<int>, greater<int>> minQ;
};

/**
 * Your MedianFinder object will be instantiated and called as such:
 * MedianFinder* obj = new MedianFinder();
 * obj->addNum(num);
 * double param_2 = obj->findMedian();
 */

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type MedianFinder struct {
    minq hp
    maxq hp
}

func Constructor() MedianFinder {
    return MedianFinder