239. Sliding Window Maximum

• Difficulty: Hard

• Topics: Heap, Sliding Window

• Similar Questions:

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Problem:

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.

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

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

Follow up:
Could you solve it in linear time?

Solutions:

class Solution {
public:
    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
        deque<int> q;
        for (int i = 0; i < k - 1; ++i) {
            while (!q.empty() && q.back() < nums[i]) {
                q.pop_back();
            }
            q.push_back(nums[i]);
        }

        vector<int> ret;
        for (int i = k - 1; i < nums.size(); ++i) {
            int val = nums[i];
            while (!q.empty() && q.back() < val) {
                q.pop_back();
            }
            q.push_back(val);
            ret.push_back(q.front());
            if (nums[i - (k - 1)] == q.front()) {
                q.pop_front();
            }
        }

        return ret;
    }
};