Leetcode #295: Find Median from Data Stream
In this guide, we solve Leetcode #295 Find Median from Data Stream in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Design, Two Pointers, Data Stream, Sorting, Heap (Priority Queue)
Intuition
The constraints hint that we can reason about two ends of the data at once, which is perfect for a two-pointer scan.
Moving one pointer at a time keeps the invariant intact and avoids nested loops.
Approach
Place pointers at the left and right ends and move them based on the comparison or target condition.
This yields a clean linear pass after any required sorting.
Steps:
- Set left and right pointers.
- Move a pointer based on the condition.
- Update the best answer while scanning.
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
Python Solution
class MedianFinder:
def __init__(self):
self.minq = []
self.maxq = []
def addNum(self, num: int) -> None:
heappush(self.minq, -heappushpop(self.maxq, -num))
if len(self.minq) - len(self.maxq) > 1:
heappush(self.maxq, -heappop(self.minq))
def findMedian(self) -> float:
if len(self.minq) == len(self.maxq):
return (self.minq[0] - self.maxq[0]) / 2
return self.minq[0]
# Your MedianFinder object will be instantiated and called as such:
# obj = MedianFinder()
# obj.addNum(num)
# param_2 = obj.findMedian()
Complexity
The time complexity is . The space complexity is , where is the number of elements.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.