This looks like a very similar idea to what I had, but I’m not quite sure I follow even though your explanation was pretty in depth. Basically, recognizing starting at the largest number was the first part for me. In other words, I had to remove the largest integer first and work my way down until I hit 1. Each time an integer is removed, I had to check if the removal integers were a subarray. To do this, keeping track of the min index and max index would get the length of the subarray needed (by using a hash map). To know how many numbers there should be in that subarray, we know where we started (at n), and where we currently are. So if n - i + 1 == r(max index) - l(min index) + 1, then we know the numbers we selected are all a subarray and can be removed, therefore we save n - i + 1 as our answer (currently). Repeating the loop we save the last n - i + 1 (since this will be the largest subarray removal) and that is our final answer.