Question: A long array A[] is given to you. There is a sliding window of size w, which is moving from the very left of the array to the very right. You can only see the w numbers in the window. Each time the sliding window moves rightwards by one position.

Example: The array is [1 3 -1 -3 5 3 6 7], and w is 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```

Input: A long array A[], and a window width w
Output: An array B[], B[i] is the maximum value of from A[i] to A[i+w-1]

Answer: The obvious solution with run time complexity of O(nw) is definitely not efficient enough. Every time the window is moved, we have to search for the maximum from w elements in the window.

We need a data structure where we can store the candidates for maximum value in the window and discard the element, which are outside the boundary of window. For this, we need a data structure in which we can edit at both the ends, front and back. Deque is a perfect candidate for this problem.

We are trying to find a way in which, we need not search for maximum element among all in the window. We will make sure that the largest element in the window would always appear in the front of the queue.
While traversing the array in forward direction if we find a window where element A[i] > A[j] and i > j, we can surely say that A[j], will not be the maximum element for this and succeeding windows. So there is no need of storing j in the queue and we can discard A[j] forever.
For example, if the current queue has the elements: [4 13 9], and a new element in the window has the element 15. Now, we can empty the queue without considering elements 4, 13, and 9, and insert only element 15 into the queue.

Every time, we move to a new window, we will be getting a new element and leave an old element. We should take care of:
1. Popping elements outside the window from queue front.
2. Popping elements that are less than new element from the queue.
3. Push new element in the queue as per above discussion.
At each iteration the queue will have indexes of following elements:
```Window position               Queue content        Max
---------------               -------------       ----- ```
```[1  3  -1] -3  5  3  6  7       3 -1                3
1  [3  -1  -3] 5  3  6  7       3 -1 -3             3
1  3  [-1  -3  5] 3  6  7       5                   5
1  3  -1  [-3  5  3] 6  7       5 3                 5
1  3  -1  -3  [5  3  6] 7       6                   6
1  3  -1  -3  5  [3  6  7]      7                   7```

Code:
```
#include<deque>

void maxSlidingWindow(int A[], int n, int w, int B[])
{
deque<int> Q;

//Initilize deque Q for first window
for (int i = 0; i < w; i++)
{
while (!Q.empty() && A[i] >= A[Q.back()])
Q.pop_back();
Q.push_back(i);
}

for (int i = w; i < n; i++)
{
B[i-w] = A[Q.front()];

//update Q for new window
while (!Q.empty() && A[i] >= A[Q.back()])
Q.pop_back();

//Pop older element outside window from Q
while (!Q.empty() && Q.front() <= i-w)
Q.pop_front();

//Insert current element in Q
Q.push_back(i);
}
B[n-w] = A[Q.front()];
}```

Complexity: Each element in the list is being inserted and then removed at most once. Therefore, the total number of insert and delete operations is 2n. Therefore it is an O(n) solution.