leetcode - 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:

-10^5 <= num <= 10^5
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?

Solution

Solved after help.

Use 2 heaps to store all the numbers, large heap store all the large half of numbers, and small stores all the small half of numbers. large is a min-root heap, and small is a big-root heap. When we need to calculate the median, just find the top element of each heap.

To maintain their property, every time there’s a new number, push it into one heap, and then pop from the heap to another. For example, let’s say we restrain the size of small to be n//2, then when their sizes are the same, the next new element goes to large.

When adding the element, firstly add the element into small, and pop one element from small to add into large, by doing so we could make sure all the elements in large are larger than small.
Similarly, when adding an element into small, firstly add the element into large, then pop one element from large and add it to small.

Time complexity: $o(n\log n)$ for adding, $o(1)$ for get median.
Space complexity: $o(n)$

Code

class MedianFinder:

    def __init__(self):
        self.small = []
        self.large = []


    def addNum(self, num: int) -> None:
        if len(self.small) == len(self.large):
            heapq.heappush(self.large, -heapq.heappushpop(self.small, -num))
        else:
            heapq.heappush(self.small, -heapq.heappushpop(self.large, num))
        

    def findMedian(self) -> float:
        if len(self.small) == len(self.large):
            return (self.large[0] - self.small[0]) / 2
        else:
            return self.large[0]
        


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