Leetcode #480: Sliding Window Median
In this guide, we solve Leetcode #480 Sliding Window Median 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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Array, Hash Table, Sliding Window, Heap (Priority Queue)
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input: nums = [1,3,-1,-3,5,3,6,7], k = 3
Output: [1.00000,-1.00000,-1.00000,3.00000,5.00000,6.00000]
Explanation:
Window position Median
--------------- -----
[1 3 -1] -3 5 3 6 7 1
1 [3 -1 -3] 5 3 6 7 -1
1 3 [-1 -3 5] 3 6 7 -1
1 3 -1 [-3 5 3] 6 7 3
1 3 -1 -3 [5 3 6] 7 5
1 3 -1 -3 5 [3 6 7] 6
Python Solution
class MedianFinder:
def __init__(self, k: int):
self.k = k
self.small = []
self.large = []
self.delayed = defaultdict(int)
self.small_size = 0
self.large_size = 0
def add_num(self, num: int):
if not self.small or num <= -self.small[0]:
heappush(self.small, -num)
self.small_size += 1
else:
heappush(self.large, num)
self.large_size += 1
self.rebalance()
def find_median(self) -> float:
return -self.small[0] if self.k & 1 else (-self.small[0] + self.large[0]) / 2
def remove_num(self, num: int):
self.delayed[num] += 1
if num <= -self.small[0]:
self.small_size -= 1
if num == -self.small[0]:
self.prune(self.small)
else:
self.large_size -= 1
if num == self.large[0]:
self.prune(self.large)
self.rebalance()
def prune(self, pq: List[int]):
sign = -1 if pq is self.small else 1
while pq and sign * pq[0] in self.delayed:
self.delayed[sign * pq[0]] -= 1
if self.delayed[sign * pq[0]] == 0:
self.delayed.pop(sign * pq[0])
heappop(pq)
def rebalance(self):
if self.small_size > self.large_size + 1:
heappush(self.large, -heappop(self.small))
self.small_size -= 1
self.large_size += 1
self.prune(self.small)
elif self.small_size < self.large_size:
heappush(self.small, -heappop(self.large))
self.large_size -= 1
self.small_size += 1
self.prune(self.large)
class Solution:
def medianSlidingWindow(self, nums: List[int], k: int) -> List[float]:
finder = MedianFinder(k)
for x in nums[:k]:
finder.add_num(x)
ans = [finder.find_median()]
for i in range(k, len(nums)):
finder.add_num(nums[i])
finder.remove_num(nums[i - k])
ans.append(finder.find_median())
return ans
Complexity
The time complexity is , and the space complexity is . The space complexity is .
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.