LeetCode: Find Median from Data Stream

Posted on April 3, 2018February 14, 2020 by braindenny

Find Median from Data Stream

Similar Problems:

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

Examples:

[2,3,4] , the median is 3

[2,3], the median is (2 + 3) / 2 = 2.5

Design a data structure that supports the following two operations:

void addNum(int num) – Add a integer number from the data stream to the data structure.
double findMedian() – Return the median of all elements so far.
For example:

addNum(1)
addNum(2)
findMedian() -> 1.5
addNum(3) 
findMedian() -> 2

Follow up:

If all integer numbers from the stream are between 0 and 100, how would you optimize it?
If 99% of all integer numbers from the stream are between 0 and 100, how would you optimize it?

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

## https://code.dennyzhang.com/find-median-from-data-stream
## Basic Ideas: Use 2 heaps: 
##      maxheap for the first half, minheap for the second half
##
##  Notice: When taking new values, we can't appeneding them directly.
##          It will not be an ordred integer list then.
##          What if we have duplicate values
##
## Complexity: Time O(log(n)), Space O(n)
import heapq
class MedianFinder:

    def __init__(self):
        """
        initialize your data structure here.
        """
        self.left_q, self.right_q = [], []
        heapq.heapify(self.left_q)
        heapq.heapify(self.right_q)

    def addNum(self, num):
        """
        :type num: int
        :rtype: void
        """
        if len(self.left_q) == 0:
            heapq.heappush(self.left_q, -num)
            return

        if num <= -self.left_q[0]:
            # should insert to left
            heapq.heappush(self.left_q, -num)
            # rebalancing
            if len(self.left_q) > len(self.right_q)+1:
                element = -heapq.heappop(self.left_q)
                heapq.heappush(self.right_q, element)
        else:
            heapq.heappush(self.right_q, num)
            # rebalancing
            if len(self.right_q) > len(self.left_q):
                new_element = -heapq.heappop(self.right_q)
                heapq.heappush(self.left_q, new_element)

    def findMedian(self):
        """
        :rtype: float
        """
        if len(self.left_q) == 0: return None
        if (len(self.left_q) == len(self.right_q)):
            return (-self.left_q[0] + self.right_q[0])/2
        else:
            return float(-self.left_q[0])

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

Post Views: 4