[剑指Offer] Day 27: Stack and Queue

Authored by Tony Feng

Created on May 8th, 2022

Last Modified on May 8th, 2022

Task 1 - Q59,I. 滑动窗口的最大值

Question

给定一个数组 nums 和滑动窗口的大小 k，请找出所有滑动窗口里的最大值。你可以假设 k 总是有效的，在输入数组不为空的情况下，1 ≤ k ≤ 输入数组的大小。

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输入: nums = [1,3,-1,-3,5,3,6,7], 和 k = 3
输出: [3,3,5,5,6,7] 
解释: 

  滑动窗口的位置                最大值
---------------               -----
[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

Solution

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class Solution:
    def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]:
        ans, window = [], [] # Record the indices whose nums are in descending order
        for i in range(len(nums)):
            # Remove the index whose elements < the new num
            while window and nums[i] > nums[window[-1]]:
                window.pop()

            # Add the index of the new num
            window.append(i)

            # if the size of the window > k, pop the first index in the window
            if i - window[0] + 1 > k:
                window.pop(0)

            # Add the first element after the window is formed
            if i >= k - 1:
                ans.append(nums[window[0]])
        return ans

Explanation

Time Complexity: O(N)
Space Complexity: O(K), i.e. It depends on the k

Task 2 - Q59,II. 队列的最大值

Question

请定义一个队列并实现函数 max_value 得到队列里的最大值，要求函数max_value、push_back 和 pop_front 的均摊时间复杂度都是O(1)。若队列为空，pop_front 和 max_value 需要返回 -1

Solution

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import queue
class MaxQueue: 
    def __init__(self):
        self.queue = queue.Queue()
        self.deque = queue.deque() # Double-ended queue, the left is always the largest

    def max_value(self) -> int:
        return self.deque[0] if self.deque else -1

    def push_back(self, value: int) -> None:
        self.queue.put(value)
        while self.deque and self.deque[-1] < value:
            self.deque.pop()
        self.deque.append(value)

    def pop_front(self) -> int:
        if self.queue.empty(): 
            return -1
        val = self.queue.get()
        if val == self.deque[0]:
            self.deque.popleft()
        return val

Explanation

Time Complexity
- max_value: O(1)
- pop_front: O(1)
- push_back: O(1)
  - For example, 543216，the last push_back takes O(N) and the rest of each only needs O(1). So the average is (O(1)*(N-1)+O(N))/N=O(1).
Space Complexity: O(N)