using deque to maintain maximum elements of a window

Author: Arjiit

Deque/maxSlidingWindow.java

@@ -0,0 +1,35 @@
+class Solution {
+    public int[] maxSlidingWindow(int[] nums, int k) {
+       /*
+       We use Double ended queue to maintain the greatest element at the front 
+       of the queue. And smaller elements behind it.
+       If the element is not in the sliding window, we remove it from the queue.
+       1. We check if the element at the front of deque is in the window or not.
+       2. We check if the element at the end of deque is less than current element.
+       3. we add the current element to the end of the deque to maintain the order.
+       */
+    
+        if (nums == null || nums.length == 0 ||nums.length < k) return new int[0];
+        
+        Deque<Integer> dq = new ArrayDeque<>();
+        int[] res = new int[nums.length - k + 1];
+        int i = 0;
+        while(i < nums.length) { 
+            // from the front of deque
+            if (!dq.isEmpty() && dq.peekFirst() == i - k) { // storing indexes
+                dq.pollFirst();
+            }
+            // from the end of deque, keep removing till element is smaller than current element
+            while(!dq.isEmpty() && nums[dq.peekLast()] < nums[i]) {
+                dq.pollLast();
+            }
+            dq.offerLast(i);
+            
+            if (i >= k-1) {
+                res[i-k+1] = nums[dq.peekFirst()]; // max element will be at top
+            }
+            i++;
+        }
+        return res;
+    }
+}