**Difficulty: Hard**

**Frequency: N/A**

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.

For example,

Given *nums* = `[1,3,-1,-3,5,3,6,7]`

, and *k* = 3.

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

Therefore, return the max sliding window as `[3,3,5,5,6,7]`

.

**Note: **

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

**Follow up:**

Could you solve it in linear time?

**Hint:**

- How about using a data structure such as deque (double-ended queue)?
- The queue size need not be the same as the window’s size.
- Remove redundant elements and the queue should store only elements that need to be considered.

**Test Cases:**

**Solution 1:**

**Data structure:**

**Steps:**

**Complexity:**

Runtime:

Space:

**Code:**

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

**Solution 2:**

**Data structure:**

**steps:**

**Complexity**:

Runtime:

Space:

**Code**:

**Submission errors:**

**Things to learn:**

Advertisements