 Solution

Store the current multiset in an unsorted prefix a[0..len); Add appends exactly as in part (a). The coupling invariant is equality of length and multiset with abstract xs, rather than equality of insertion order. Get applies quickselect:The loop invariant is $0\le l\le k<r\le \mathrm{len}$, the complete array has its original multiset, every discarded entry on the left is at most every active entry, and every active entry is at most every discarded entry on the right. Thus the desired global order statistic is the local one of rank $k-l$. The two-way array partition result either identifies it or excludes the pivot and the irrelevant side. The positive loop variant $r-l$ strictly decreases unless the procedure returns.

Balanced pivots give $T(n)\le T(\lceil n/2\rceil )+cn$, whose geometric series is $O(n)$. More generally, choosing a uniformly random pivot before each call, by swapping a random active entry into a[r-1], gives expected $\Theta (n)$ running time on distinct keys: with constant probability, the pivot is in the middle half and the retained side has size at most $3n/4$. Expected work over these shrinking stages is a convergent geometric series.The particular last-entry pivot in the displayed code has no unconditional expected guarantee without an input distribution. Its worst case repeatedly discards one entry. Duplicate keys require further care: on an all-equal segment, the strict partition always returns $l$, so selecting the last rank costs $n+(n-1)+\cdots +1=\Theta (n^2)$ even with randomized pivot positions. A three-way array partitioning that groups all keys equal to the pivot restores the usual randomized linear expectation with duplicates; alternatively the stated expectation assumes distinct keys. This is a qualification to the bound, not to the algorithm correctness of the displayed code.