239. Sliding Window Maximum

Intuition

The problem requires us to find the maximum value in each sliding window of size k as it moves from the left to the right of the array. A monotonic queue is an efficient data structure that can help keep track of maximum values in a sliding window, maintaining the elements in a way that allows us to access the maximum value quickly.

Approach: Monotonic Queue

Explanation:

1. Initialization:
2. We use a deque dq to store the elements of the current window in a way that the front of the deque always contains the maximum value of the current window.
3. Sliding Window Process:
4. As we iterate through the array, we manage the deque to maintain the properties of the sliding window and ensure it remains monotonic.
5. Maintaining the Deque:
6. For each element nums[i]:
   • Remove elements from the back of the deque while the current element nums[i] is greater than the elements at the back. This ensures that the deque only contains potential candidates for the maximum value.
   • Add the current element to the back of the deque.
7. Sliding the Window:
8. When the window is fully formed (i.e., when i - k + 1 >= 0):
   • If the element that is going out of the window (i.e., nums[i - k + 1]) is the same as the element at the front of the deque, remove it from the front. This ensures that we only keep elements within the current window in the deque.
   • Store the current maximum value (i.e., dq.front()) in the array nums at the position corresponding to the start of the window.
9. Final Adjustments:
10. Resize the array nums to contain only the results of the maximum values for each window.

Complexity

• Time complexity: O(n)
• Space complexity: O(k)

Code

class Solution {
public:
    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
        deque<int> dq;

        for (int i = 0; i < (int)nums.size(); i++) {
            while (!dq.empty() && nums[i] > dq.back()) {
                dq.pop_back();
            }

            dq.push_back(nums[i]);

            if (i - k + 1 >= 0) {
                if (nums[i - k + 1] == dq.front()) {
                    dq.pop_front();
                }
                else {
                    nums[i - k + 1] = dq.front();
                }
            }
        }

        nums.resize(nums.size() - k + 1);

        return move(nums);
    }
};