Sliding Window Maximum (Deque)

/*
 * Problem : Window maxima.
 * Input   : n, array, k.
 * Output  : Maxima.
 * Approach: Deque.
 * Time    : O(n)
 * Space   : O(k)
 * Java    : 8 or newer.
 */
import java.util.Scanner;
import java.util.Arrays;
import java.util.stream.Collectors;

public class Main {
    // This method solves the task. It receives the input and returns the answer.
    static int[] solve(int[] a, int k) {
        // Step 1: The deque holds indexes. Their values go down from front to back, so the front is the maximum.
        java.util.ArrayDeque<Integer> deque = new java.util.ArrayDeque<>();
        int[] result = new int[a.length - k + 1];
        for (int i = 0; i < a.length; i++) {
            // Step 2: Remove the front index when it has left the window.
            if (!deque.isEmpty() && deque.peekFirst() <= i - k) {
                deque.pollFirst();
            }
            // Step 3: Remove smaller values from the back. They can never be a maximum again.
            while (!deque.isEmpty() && a[deque.peekLast()] <= a[i]) {
                deque.pollLast();
            }
            deque.addLast(i);
            // Step 4: When the first window is full, the front is the answer for this window.
            if (i >= k - 1) {
                result[i - k + 1] = a[deque.peekFirst()];
            }
        }
        return result;
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int size = scanner.nextInt();
        int[] a = new int[size];
        for (int i = 0; i < size; i++) {
            a[i] = scanner.nextInt();
        }
        int k = scanner.nextInt();
        System.out.println(Arrays.stream(solve(a, k)).mapToObj(String::valueOf).collect(Collectors.joining(" ")));
    }
}