## Problem Description

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?

## Solution

• C++
vector<int> maxSlidingWindow(vector<int> & nums, int k) {
// A container for index in the current window
deque<int> dq;
// A container for max values as a result
vector<int> result;
// Iterate from begin to end in the array
for (int i = 0; i<nums.size(); i++) {
// Erase value in the front as the window move forward
if (!dq.empty() && dq.front() == i-k) dq.pop_front();
// Keep only the index of the largest number in the window and
// the index of the rightmost number in the window in container dq
while (!dq.empty() && nums[dq.back()] < nums[i]) dq.pop_back();
dq.push_back(i);
// Push the max value to the result vector
if (i >= k-1) result.push_back(nums[dq.front()]);
}
return result;
}