solution.en.md

Solution

Contributions: Codex (GPT-6) specified the stack rules and supplied the visibility segment tree, block alternative, proofs, implementations, and verification.

Strict maxima from the right

An element remains in the final stack exactly when no later element is at least as large. Such a later element eventually pops it; without one, no later insertion can pop it. Thus the answer is the number of strict record maxima when scanning from right to left.

Store each segment-tree node's maximum $M$ and unrestricted record count $C$ . Let $F(u,t)$ count its right-to-left records greater than threshold $t$ . If $M_u\le t$ , return zero. A leaf otherwise contributes one.

For children $L,R$ , the right interval is scanned first, leaving threshold $M_R$ for the left interval. Therefore

$$C_u=C_R+F(L,M_R).$$

To evaluate $F(u,t)$ :

• If $M_R\le t$ , every right element is blocked. Recurse only into $F(L,t)$ .
• If $M_R>t$ , the right maximum survives and fixes the left threshold at $M_R$ . The left contribution is already stored as $C_u-C_R$ . Recurse only into $F(R,t)$ and add that difference.

Only one child is visited per level, so $F$ takes $O(\log n)$ time.

Updates and interval queries

After a point update, recompute maxima and record counts toward the root. Each merge makes one $F$ call, giving $O(\log^2 n)$ time.

Process the canonical nodes of a query from right to left. Start the threshold at $-1$ . For a covered node, add $F(u,t)$ and replace the threshold by $\max(t,M_u)$ . This exactly represents the restriction imposed by the already scanned suffix. There are $O(\log n)$ covered nodes, so the query costs $O(\log^2 n)$ .

The record equivalence, merge identity, and threshold invariant prove that each answer equals the specified stack size. Equality must block a left element, consistent with strict record maxima.

Complexity and alternatives

Let $n$ be the sequence length and $m$ the operation count. A node of height $h$ needs $O(h)$ construction work, and there are $O(n/2^h)$ such nodes. Summing gives $O(n)$ construction. Each update or query takes $O(\log^2 n)$ time; space is $O(n)$ .

std.cpp is the preferred implementation. The block alternative updates in $O(B)$ and queries in $O(B+(n/B)\log B)$ . It has a simpler structure, uses linear space, and also passes the complete data.

Measurements

Each formal input has five timing runs and a separate peak-memory run. The standard program's slowest median is 0.055566 seconds and peak memory is 6.90 MiB. Twice the median, rounded up to 0.5 seconds, gives a 0.5-second limit. The alternative's slowest median is 0.063319 seconds and peak memory is 5.62 MiB. See the environment and individual measurements. Recalibrate in other environments.