Faster merge with less auxiliary space?

In Algorithms fourth edition by Robert Sedgewick and Kevin Wayne, the exercise 2.2.10 states that:

Implement a version of merge() that copies the second half of a[] to aux[] in decreasing order and then does the merge back to a[]. This change al- lows you to remove the code to test that each of the halves has been exhausted from the inner loop.

This is my code in Python:

def merge(a, lo, mi, hi):      aux_hi = deque(a[mi:hi])     for i in range(lo, hi)[::-1]:         if aux_hi:              last_a = i-len(aux_hi)             if last_a < lo:                  a[i] = aux_hi.pop()             else:                  a[i] = aux_hi.pop() if aux_hi[-1] > a[last_a] else a[last_a]         else:               break 

Have I implemented it as required? Or can it be improved further?