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Вот решение в O (n) временной сложности с вспомогательным deque

public class TestSlidingWindow {

    public static void main(String[] args) {
        int[] arr = { 1, 5, 7, 2, 1, 3, 4 };
        int k = 3;

        printMaxInSlidingWindow(arr, k);

    }

    public static void printMaxInSlidingWindow(int[] arr, int k) {

        Deque<Integer> queue = new ArrayDeque<Integer>();
        Deque<Integer> auxQueue = new ArrayDeque<Integer>();

        int[] resultArr = new int[(arr.length - k) + 1];

        int maxElement = 0;
        int j = 0;
        for (int i = 0; i < arr.length; i++) {

            queue.add(arr[i]);

            if (arr[i] > maxElement) {
                maxElement = arr[i];
            }

    /** we need to maintain the auxiliary deque to maintain max element  in case max element is removed. 
        We add the element to deque straight away if subsequent element is less than the last element
        (as there is a probability if last element is removed this element can be max element) otherwise 
        remove all lesser element then insert current element **/

            if (auxQueue.size() > 0) {
                if (arr[i] <  auxQueue.peek()) {
                    auxQueue.push(arr[i]);
                } else {
                    while (auxQueue.size() > 0 && (arr[i] >  auxQueue.peek())) {
                        auxQueue.pollLast();
                    }

                    auxQueue.push(arr[i]);
                }
            }else {
                auxQueue.push(arr[i]);
            }

            if (queue.size() > 3) {
                int removedEl = queue.removeFirst();
                if (maxElement == removedEl) {
                    maxElement = auxQueue.pollFirst();
                }
            }

            if (queue.size() == 3) {
                resultArr[j++] = maxElement;
            }

        }

        for (int i = 0; i < resultArr.length; i++) {
            System.out.println(resultArr[i]);
        }
    }

}