|
| 1 | +""" |
| 2 | +Problem Link: https://leetcode.com/problems/find-median-from-data-stream/ |
| 3 | + |
| 4 | +The median is the middle value in an ordered integer list. If the size of the list is even, |
| 5 | +there is no middle value and the median is the mean of the two middle values. |
| 6 | +For example, for arr = [2,3,4], the median is 3. |
| 7 | +For example, for arr = [2,3], the median is (2 + 3) / 2 = 2.5. |
| 8 | + |
| 9 | +Implement the MedianFinder class: |
| 10 | +MedianFinder() initializes the MedianFinder object. |
| 11 | +void addNum(int num) adds the integer num from the data stream to the data structure. |
| 12 | +double findMedian() returns the median of all elements so far. Answers within 10-5 of the actual |
| 13 | +answer will be accepted. |
| 14 | + |
| 15 | +Example 1: |
| 16 | +Input |
| 17 | +["MedianFinder", "addNum", "addNum", "findMedian", "addNum", "findMedian"] |
| 18 | +[[], [1], [2], [], [3], []] |
| 19 | +Output |
| 20 | +[null, null, null, 1.5, null, 2.0] |
| 21 | + |
| 22 | +Explanation |
| 23 | +MedianFinder medianFinder = new MedianFinder(); |
| 24 | +medianFinder.addNum(1); // arr = [1] |
| 25 | +medianFinder.addNum(2); // arr = [1, 2] |
| 26 | +medianFinder.findMedian(); // return 1.5 (i.e., (1 + 2) / 2) |
| 27 | +medianFinder.addNum(3); // arr[1, 2, 3] |
| 28 | +medianFinder.findMedian(); // return 2.0 |
| 29 | + |
| 30 | +Constraints: |
| 31 | +-105 <= num <= 105 |
| 32 | +There will be at least one element in the data structure before calling findMedian. |
| 33 | +At most 5 * 104 calls will be made to addNum and findMedian. |
| 34 | + |
| 35 | +Follow up: |
| 36 | +If all integer numbers from the stream are in the range [0, 100], how would you optimize your solution? |
| 37 | +If 99% of all integer numbers from the stream are in the range [0, 100], how would you optimize your |
| 38 | +solution? |
| 39 | +""" |
| 40 | +import heapq |
| 41 | + |
| 42 | + |
| 43 | +class MedianFinder: |
| 44 | + |
| 45 | + def __init__(self): |
| 46 | + self.min_heap = [] |
| 47 | + self.max_heap = [] |
| 48 | + |
| 49 | + |
| 50 | + # Time Complexity: O(logn) |
| 51 | + def addNum(self, num: int) -> None: |
| 52 | + heapq.heappush(self.max_heap, -num) |
| 53 | + heapq.heappush(self.min_heap, -heapq.heappop(self.max_heap)) |
| 54 | + if len(self.min_heap) > len(self.max_heap): |
| 55 | + heapq.heappush(self.max_heap, -heapq.heappop(self.min_heap)) |
| 56 | + |
| 57 | + |
| 58 | + # Time Complexity: O(1) |
| 59 | + def findMedian(self) -> float: |
| 60 | + if len(self.max_heap) > len(self.min_heap): |
| 61 | + return -self.max_heap[0] |
| 62 | + |
| 63 | + return (-self.max_heap[0] + self.min_heap[0]) / 2 |
| 64 | + |
| 65 | + |
| 66 | + |
| 67 | +# Your MedianFinder object will be instantiated and called as such: |
| 68 | +# obj = MedianFinder() |
| 69 | +# obj.addNum(num) |
| 70 | +# param_2 = obj.findMedian() |
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