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# Sliding Window Maximum

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`

```java
// Other way to do this is using leftmax and rightmax arrays with boxed partitions
// https://drive.google.com/file/d/14k1lhaQe9He5EouLXuSlXrMR61HUBMy1/view?usp=sharing
class Solution {
    public int[] maxSlidingWindow(int[] nums, int k) {
        Deque<Integer> dq = new LinkedList<>();
        int[] ans = new int[nums.length - k + 1];
        int index = 0;
        for (int i = 0; i < nums.length; i++) {
            // Removing the elements our of range 'k'
            while (dq.size() != 0 && dq.peekFirst() < i - (k - 1))
                dq.pollFirst();
            // Maintaining increasing deque from right -> left
            while (dq.size() != 0 && nums[dq.peekLast()] < nums[i])
                dq.pollLast();
            dq.addLast(i);
            // If our window of size 'k' is ready
            if (i >= k - 1) {
                ans[index++] = nums[dq.peekFirst()];
            }
        }
        return ans;
    }
}
```