239. Sliding Window Maximum

239. Sliding Window Maximum (Hard)

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. Return the max sliding window.

Follow up:
Could you solve it in linear time?

Example:

Input: nums = [1,3,-1,-3,5,3,6,7], and k = 3
Output: [3,3,5,5,6,7] 
Explanation: 

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

 

Constraints:

• 1 <= nums.length <= 10^5
• -10^4 <= nums[i] <= 10^4
• 1 <= k <= nums.length

Related Topics:
Heap, Sliding Window

Similar Questions:

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Solution 1. Monotonous Deque

Assume the array is [3, 1, 2, ...] and k = 3, popping 3 out of the window will result in max value update, but popping 1 won't. This means that we can just keep track of [3, 2], i.e. a monotonically decreasing sequence of values.

Here we store the index of the monotonoically decreasing sequence. When a new value A[i] is added to the window, we pop the trailing index in the deque which are pointing to values that are smaller or equal to A[i]. Then we can push i into the deque.

We need to pop the index which falls out of the window from the deque as well.

// OJ: https://leetcode.com/problems/sliding-window-maximum/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
    vector<int> maxSlidingWindow(vector<int>& A, int k) {
        vector<int> ans;
        deque<int> q;
        for (int i = 0; i < A.size(); ++i) {
            if (q.size() && q.front() == i - k) q.pop_front();
            while (q.size() && A[q.back()] <= A[i]) q.pop_back();
            q.push_back(i);
            if (i >= k - 1) ans.push_back(A[q.front()]);
        }
        return ans;
    }
};

Solution 2. Monotonous Deque

Similar to Solution 1, but here we store the values instead of the indexes in the deque.

// OJ: https://leetcode.com/problems/sliding-window-maximum/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
    vector<int> maxSlidingWindow(vector<int>& A, int k) {
        deque<int> q;
        vector<int> ans;
        for (int i = 0; i < A.size(); ++i) {
            if (i >= k && q.size() && q.front() == A[i - k]) q.pop_front();
            while (q.size() && q.back() < A[i]) q.pop_back();
            q.push_back(A[i]);
            if (i >= k - 1) ans.push_back(q.front());
        }
        return ans;
    }
};