Sliding Window Maximum
Given an array and sliding window of size k, return the maximum in each window position.
Output space-separated.
Example:
1,3,-1,-3,5,3,6,7 3
3 3 5 5 6 7
- The array is processed with a sliding window of size k=3, starting from the first element.
- For each window position, the maximum value is found:
- Window 1: max(1,3,−1)=3
- Window 2: max(3,−1,−3)=3
- Window 3: max(−1,−3,5)=5
- Window 4: max(−3,5,3)=5
- Window 5: max(5,3,6)=6
- Window 6: max(3,6,7)=7
- The maximum values from each window are collected and output as space-separated values.
- The final output is therefore: 335567
Constraints:
- 1 <= len(nums) <= 10^5
- -10^4 <= nums[i] <= 10^4
- 1 <= k <= len(nums)
Background Knowledge
The Sliding Window Maximum problem involves finding the maximum element in each window of a given size k as it slides over an array. This problem is a classic example of a window-based problem, where we need to process a subset of the data at a time. To solve this problem, we need to understand the concept of queues and deques (double-ended queues), which are data structures that allow efficient insertion and removal of elements from both ends. We also need to understand the concept of priority queues, which are data structures that allow efficient insertion and removal of elements based on their priority.
The key concept in this problem is the use of a deque to keep track of the indices of the elements in the current window. We can use the deque to efficiently remove elements that are out of the current window and add new elements that enter the window. We also need to understand the concept of maximum and how to find it in a window of elements. This can be done using a priority queue, which can keep track of the maximum element in the window.
The Sliding Window Maximum problem is a variation of the Maximum Subarray problem, which involves finding the maximum sum of a subarray of a given size. However, in this problem, we need to find the maximum element in each window, rather than the maximum sum. This requires a different approach, using a deque to keep track of the indices of the elements in the current window, and a priority queue to keep track of the maximum element in the window.
Algorithm/Approach
The general approach to solve this type of problem is to use a deque to keep track of the indices of the elements in the current window, and a priority queue to keep track of the maximum element in the window. We can iterate over the array, adding elements to the deque and priority queue as we go, and removing elements that are out of the current window. We can use the priority queue to efficiently find the maximum element in the window at each position.
Step-by-Step Strategy
To implement the solution, we can follow these steps:
- Initialize an empty deque to keep track of the indices of the elements in the current window
- Initialize an empty priority queue to keep track of the maximum element in the window
- Iterate over the array, adding elements to the deque and priority queue as we go
- For each element, check if it is within the current window, and if not, remove it from the deque and priority queue
- Use the priority queue to find the maximum element in the window at each position
- Output the maximum element at each position
Common Pitfalls
Some common pitfalls to watch out for when implementing this solution include:
- Failing to remove elements from the deque and priority queue that are out of the current window
- Failing to update the priority queue when a new element enters the window
- Using an inefficient data structure, such as a list or array, to keep track of the elements in the window
Time & Space Complexity
The expected time complexity of this solution is O(n), where n is the length of the array, since we only need to iterate over the array once. The expected space complexity is O(k), where k is the size of the window, since we need to store the indices of the elements in the current window in the deque. The priority queue will also require O(k) space to store the maximum element in the window.