Find Median from Data Stream

Author: anantkaushik

295-find-median-from-data-stream.py

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+"""
+Problem Link: https://leetcode.com/problems/find-median-from-data-stream/
+
+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?
+"""
+import heapq
+
+
+class MedianFinder:
+
+ def __init__(self):
+ self.min_heap = []
+ self.max_heap = []
+ 
+ 
+ # Time Complexity: O(logn)
+ def addNum(self, num: int) -> None:
+ heapq.heappush(self.max_heap, -num) 
+ heapq.heappush(self.min_heap, -heapq.heappop(self.max_heap))
+ if len(self.min_heap) > len(self.max_heap):
+ heapq.heappush(self.max_heap, -heapq.heappop(self.min_heap))
+
+
+ # Time Complexity: O(1)
+ def findMedian(self) -> float:
+ if len(self.max_heap) > len(self.min_heap):
+ return -self.max_heap[0]
+ 
+ return (-self.max_heap[0] + self.min_heap[0]) / 2
+ 
+
+
+# Your MedianFinder object will be instantiated and called as such:
+# obj = MedianFinder()
+# obj.addNum(num)
+# param_2 = obj.findMedian()