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+#Sliding Window
+
+Time        Space
+Using Iteration :             O(n*k)        O(1)
+Using AVL Trees/Priority Queue:     O(n * logk)    O(k)
+Using Deque        Amortized     O(N)         O(k)
+since each element is processed (add/remove) at most twice.
+
+Eg. a = {4 , 1 , 3 , 5 , 1, 2 , 3 , 2 , 1 , 1 , 5 }  , k=3
+Ans  =  {4, 5 , 5 , 5 , 3, 3, 3, 2, 5 }
+Use a max heap, we insert the first 3 elements. And then return the top element 4.
+
+
+```
+public int[] maxSlidingWindow(int[] nums, int k) {
+    PriorityQueue<Integer> pq = new PriorityQueue<>(Collections.reverseOrder());
+    int[] res = new int[nums.length-k+1];
+    for(int i = 0; i<k; i++) {
+        pq.add(nums[i]);
+    }
+    int j=0;
+    res[j++] = pq.peek();
+    pq.remove(nums[0]);
+   
+    for(int i=k; i<nums.length;i++) {
+        pq.add(nums[i]);
+        res[j++]=pq.peek();
+        pq.remove(nums[i-k+1]);
+    }
+   
+    return res;
+}
+```