480. Sliding Window Median

Description

The median is the middle value in an ordered integer list. If the size of the list is even, there is no middle value. So the median is the mean of the two middle values.

For examples, if arr = [2,3,4], the median is 3.
For examples, if arr = [1,2,3,4], the median is (2 + 3) / 2 = 2.5.

You are given an integer array nums and an integer k. There is a sliding window of size k which is moving from the very left of the array to the very right. You can only see the k numbers in the window. Each time the sliding window moves right by one position.

Return the median array for each window in the original array. Answers within 10^-5 of the actual value will be accepted.

Example 1:

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

Example 2:

Input: nums = [1,2,3,4,2,3,1,4,2], k = 3
Output: [2.00000,3.00000,3.00000,3.00000,2.00000,3.00000,2.00000]

Constraints:

1 <= k <= nums.length <= 10⁵
-2³¹ <= nums[i] <= 2³¹ - 1

Solutions

Solution 1: Dual Priority Queues (Min-Heap and Max-Heap) + Lazy Deletion

We can use two priority queues (min-heap and max-heap) to maintain the elements in the current window. One priority queue stores the smaller half of the elements, and the other priority queue stores the larger half of the elements. This way, the median of the current window is either the average of the top elements of the two heaps or one of the top elements.

We design a class $\textit{MedianFinder}$ to maintain the elements in the current window. This class includes the following methods:

add_num(num): Adds $\textit{num}$ to the current window.
find_median(): Returns the median of the elements in the current window.
remove_num(num): Removes $\textit{num}$ from the current window.
prune(pq): If the top element of the heap is in the lazy deletion dictionary $\textit{delayed}$, it pops the top element from the heap and decrements its lazy deletion count. If the lazy deletion count of the element becomes zero, it removes the element from the lazy deletion dictionary.
rebalance(): If the number of elements in the smaller half exceeds the number of elements in the larger half by $2$, it moves the top element of the larger half to the smaller half. If the number of elements in the smaller half is less than the number of elements in the larger half, it moves the top element of the larger half to the smaller half.

In the add_num(num) method, we first consider adding $\textit{num}$ to the smaller half. If $\textit{num}$ is greater than the top element of the larger half, we add $\textit{num}$ to the larger half. Then we call the rebalance() method to ensure that the size difference between the two priority queues does not exceed $1$.

In the remove_num(num) method, we increment the lazy deletion count of $\textit{num}$. Then we compare $\textit{num}$ with the top element of the smaller half. If $\textit{num}$ is less than or equal to the top element of the smaller half, we update the size of the smaller half and call the prune() method to ensure that the top element of the smaller half is not in the lazy deletion dictionary. Otherwise, we update the size of the larger half and call the prune() method to ensure that the top element of the larger half is not in the lazy deletion dictionary.

In the find_median() method, if the current window size is odd, we return the top element of the smaller half; otherwise, we return the average of the top elements of the smaller half and the larger half.

In the prune(pq) method, if the top element of the heap is in the lazy deletion dictionary, it pops the top element from the heap and decrements its lazy deletion count. If the lazy deletion count of the element becomes zero, it removes the element from the lazy deletion dictionary.

In the rebalance() method, if the number of elements in the smaller half exceeds the number of elements in the larger half by $2$, it moves the top element of the larger half to the smaller half. If the number of elements in the smaller half is less than the number of elements in the larger half, it moves the top element of the larger half to the smaller half.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $\textit{nums}$.

Python3JavaC++Go

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