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# LC295 - Find Median from Data Stream

## Problem

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

**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

## Solution

Time complexity is about $$O(\log n)$$, and $$O(n)$$ extra space is used.

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

def addNum(self, num: int) -> None:

	# python doesn't have maxheap, most ops are log(n)
	heapq.heappush(self.small, -1 * num)
	if (self.small and self.large and (-1 * self.small[0]) > self.large[0]):
		val = -1 * heapq.heappop(self.small)
		heapq.heappush(self.large, val)
		
	if len(self.small) > len(self.large) + 1:
		val = -1 * heapq.heappop(self.small)
		heapq.heappush(self.large, val)
	
	if len(self.large) > len(self.small) + 1:
		val = heapq.heappop(self.large)
		heapq.heappush(self.small, -1 * val)

  

def findMedian(self) -> float:
	# Odd numb of elements
	if len(self.small) > len(self.large):	
		return -1 * self.small[0]

	elif len(self.large) > len(self.small):
		return self.large[0]

	return (-1 * self.small[0] + self.large[0])/2
```