An approach better than Pancake Sorting for a restricted sort

The original source of the question for this blog post is again Algo Geeks.

Here's the problem description:
There is an array in an external system which cannot be accessed directly. 
The system exposes 3 functions whose complexity is O(1) :
length() - returns the length of the array.
get(i) - returns the element at index i.
reverse(i,j) - reverses the elements in the array from index i to index j (both indexes inclusive). 

Sort the array in the best possible way using only these 3 operations? 

The direct solution to sorting in this case is a Pancake Sort (see the video embedded below).

Pancake sort has a direct implementation complexity of O(N2) as locating the max element in the unsorted section of the array is an O(N) operation. Besides this, it is obvious that one cannot use standard quicksort or heapsort directly since a swap operation
isn't supplied. However, we can simulate a swap operation using the reverse operator. It's easy to do: 

swap(i, j) = reverse(i + 1, j) reverse(i, i+1), reverse(i+1, j) where i, j are indices to the array.

Given this swap operation, we can then implement any O(n log n) sort that we desire since we already have random access via the get(i) function.

However, we can do better:

Let us define a run as a sequence of consecutive values in the array.

1. Go through the array and reverse descending runs - O(n) as reverse(i, j) is supplied. Now you have an array with many runs in ascending order. It looks something like a0, a1 ... ak, b0, b1, ... bl, c0, c1, ... cm ...
2. Perform a merge procedure pairwise for consecutive runs in the array. The merge is performed as follows:
If a0, a1, a2 ...ak b0, b1, b2... bl are consecutive runs, then:
a) Compare a0, b0 (using get() operation to get the values). 
   --- Else If b0 < a0 then perform step (b). 
   --- Else a0 is in its correct place, increment pointer to the 'a' array and repeat step 2 for a1... ak, b0 ... bl

b) reverse(a0 ... b0) and then reverse (ak ... a0) - giving b0, a0, a1 ... ak, b1, b2 .. bl. Repeat step 2 for the run a0, a1 ... ak, b1, b2 ... bl

Complexity analysis:
Step 1 can be done in O(n) time. Using 2 indices to locate boundaries of runs and an O(1) reverse call, this is easy to achieve.
Step 2 The merge procedure requires 2 get() operations (one of which may be optimized away with careful coding) and (in the worst case) 2 reverse(i,j) operations. Therefore, the cost of putting 1 element in its correct position is O(1). Hence 2 sequences of lengths n and m can be merged in O(min{n, m}) swaps and O(max{n, m}) time.

Overall complexity: 
Let us assume that we end up with K runs of lengths L1 - Lk from step 1. Cost of this step O(N)
Cost of merging adjacent runs = O(min{Li, Li+1}) (in terms of swaps), O(max{Li, Lj}) in terms of comparisons.
Total number of merges that will take place = K

The worst case for the problem will occur when all the arguments to min{ Li, Lj } are the same. This happens when the runs are of size N/K.
Therefore, cost of merge = O( N/K ) swaps (rather reverse(i,j) calls) and total number of merges = K. Hence overall swaps complexity is O( K * N/K ) = O( N ). However, overall comparison complexity remains O(N log N). So using the reverse operation, we've be able to optimize the number of swaps. However, we cannot breach the lower bound of Omega(n log n) for the number of comparisons required for sorting as this is a comparison based sort.

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