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. Following is an 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] Requirement: Find a good optimal way to get B[i].

The double-ended queue is the perfect data structure for this problem. It supports insertion/deletion from the front and back. The trick is to find a way such that the largest element in the window would always appear in the front of the queue. How would you maintain this requirement as you push and pop elements in and out of the queue?

Besides, you might notice that there are some redundant elements in the queue that we shouldn't even consider about. For example, if the current queue has the elements: [10 5 3], and a new element in the window has the element 11. Now, we could have emptied the queue without considering elements 10, 5, and 3, and insert only element 11 into the queue.

Removing redundant elements and storing only elements that need to be considered in the queue is the key to achieve the efficient O(n) solution below.

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:

- Popping elements outside the window from queue front.
- Popping elements that are less than new element from the queue.
- Push new element in the queue as per above discussion.

import java.util.ArrayDeque;

import java.util.Deque;

public class SlidingWindow {

public static void maxSlidingWindow(int A[], int n, int w, int B[]) {

Deque<Integer> Q = new ArrayDeque<Integer>();

// Initialize deque Q for first window

for (int i = 0; i < w; i++) {

while (!Q.isEmpty() && A[i] >= A[Q.getLast()])

Q.pollLast();

Q.offerLast(i);

}

for (int i = w; i < n; i++) {

B[i - w] = A[Q.getFirst()];

// update Q for new window

while (!Q.isEmpty() && A[i] >= A[Q.getLast()])

Q.pollLast();

// Pop older element outside window from Q

while (!Q.isEmpty() && Q.getFirst() <= i - w)

Q.pollFirst();

// Insert current element in Q

Q.offerLast(i);

}

B[n - w] = A[Q.getFirst()];

}

public static void main(String args[]) {

int w = 3;

int a[] = { 1, 3, -1, -3, 5, 3, 6, 7 };

int b[] = new int[a.length - w + 1];

maxSlidingWindow(a, a.length, w, b);

System.out.println("Sliding Window Maximum is ");

for (int i = 0; i < b.length; i++) {

System.out.print(b[i] + ",");

}

}

}

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.

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