Leetcode: Sliding Window Maximum

Question

Given an array nums, 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.

For example,
Given nums = [1,3,-1,-3,5,3,6,7], and k = 3.

Window position Max
————— —–
[1 3 -1] -3 5 3 6 7 3
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 5
1 3 -1 -3 [5 3 6] 7 6
1 3 -1 -3 5 [3 6 7] 7
Therefore, return the max sliding window as [3,3,5,5,6,7].

Note:
You may assume k is always valid, ie: 1 ≤ k ≤ input array’s size for non-empty array.

Follow up:
Could you solve it in linear time?

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Analysis

The naive way is to maintain a sliding window with size k. For each window, we sort it and get the max value. the time complexity is O( n * nlog(n) )

It is easy to use heap to get the maximum value. But we need to store both its value and its position. The time complexity is O( n* log(n) ).

The efficient solution takes O( n ).
Actually, if using heap, the most of elems in the bottom of heap are never be used. Think about it. Even if the toppest elem is popped, the second largest elem will be appear. The heap size is always k, some elems in the bottom are discarded.

So we use one queue. Instead of storing all elems in current sliding window, it only stores the potential large elems, decreasing subsequence. It is the same as the function of heap.

The first elem of queue is the largest one for current window. When window slides, we need to pop the first elem and add the new one.

Solution

Get idea from here1, here2

time complexity: O(n)
space complexity: O(k)

process

ind:0,  queue: [1],  ans=[]
1  3  -1 -3  5  3  6  7       

ind:1,  queue: [3],  ans=[]
1  3  -1 -3  5  3  6  7    

ind:2,  queue: [3,-1], ans=[3]
1  3  -1 -3  5  3  6  7    

ind:3,  queue: [3,-1,-3], ans=[3,3]
1  3  -1 -3  5  3  6  7 

ind:4,  queue: [5], ans=[3,3,5]
1  3  -1 -3  5  3  6  7 

ind:5,  queue: [5,3], ans=[3,3,5,5]
1  3  -1 -3  5  3  6  7 

ind:6,  queue: [6], ans=[3,3,5,5,6]
1  3  -1 -3  5  3  6  7 

ind:7,  queue: [7], ans=[3,3,5,5,6,7]
1  3  -1 -3  5  3  6  7 

Code

class Solution(object):
    def maxSlidingWindow(self, nums, k):
        """ :type nums: List[int] :type k: int :rtype: List[int] """

        res, queue = [], []

        for ind in range(len(nums)):
            # make sure the size fo range of all elems in queue is smaller than k 
            if len(queue)>0 and ind-queue[0][0]>=k:
                del queue[0]

            # we don't need to consider the small elem, which are never be the largest one in the range
            while len(queue)>0 and nums[ind] > queue[-1][1]:
                queue.pop()

            queue.append( (ind, nums[ind]) )

            # for each window, output its result
            if ind>=k-1:
                res.append( queue[0][1] )

        return res