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295. Find Median from Data Stream

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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() 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?

Craig's notes:

  • The left (max) heap contains at most one element more than the right (min) heap.
  • If the heaps are different sizes the median is at the top of the left heap.
  • If the heaps are the same sizes the median is the sum divided by 2 of the top elements of the heaps.

Solution

class MedianFinder:
    def __init__(self):
        self.left_heap = []  # max heap
        self.right_heap = []  # min heap

    def addNum(self, num: int) -> None:
        if self.left_heap == []:
            heapq.heappush(self.left_heap, -num)
            return

        m = -self.left_heap[0]
        if num <= m:
            heapq.heappush(self.left_heap, -num)
        else:
            heapq.heappush(self.right_heap, num)

        self.rebalance()

    def rebalance(self):
        if len(self.left_heap) < len(self.right_heap):
            heapq.heappush(self.left_heap, -heapq.heappop(self.right_heap))

        if len(self.left_heap) > len(self.right_heap) + 1:
            heapq.heappush(self.right_heap, -heapq.heappop(self.left_heap))

    def findMedian(self) -> float:
        if len(self.left_heap) > len(self.right_heap):
            return -self.left_heap[0]

        return (-self.left_heap[0] + self.right_heap[0]) / 2