 Solution

Assume ascending sorted arrays; descending storage can be accessed in reverse index order without copying. Among $2n$ entries, the requested rank is $k=n+1$ in ascending order, not $n$. Repeated values occupy repeated ranks.

Choose $i$ entries from the first array and $j=k-i$ from the second, using their prefixes. The feasible range is $\max (0,k-n)\le i\le \min (n,k)$. Write $A_L=A[i]$, $A_R=A[i+1]$, and similarly $B_L,B_R$ at $j$, with missing left endpoints replaced by $-\infty$ and missing right endpoints by $+\infty$. These are conceptual sentinels, not extra stored entries.

Use binary search over the feasible values of $i$:

For correctness, in the last case every left entry is no greater than every right entry, and exactly $k$ entries lie on the left. Their maximum is therefore the $k$-th smallest entry. The failed inequalities are monotone in the searched index: increasing $i$ raises the $A$ boundary and decreases the $B$ boundary, so it cannot repair $A_L>B_R$; the other case is symmetric. A partition of the first $k$ entries of a merged ordering supplies an admissible solution, including ties. Each step halves an interval with at most $n+1$ choices, giving $O(\log (n+1))$ running time and $O(1)$ auxiliary space. This is a two-sorted-array order statistic without constructing the merged array.