Direct simulation spends time on every second, even when very little changes. Instead, let $T_i$ be the time when stack $i$ becomes empty. Set $T_0=T_{n+1}=0$ for the two exterior positions.

These times are finite. The first stack loses one block every second until it is empty. Once stack $i-1$ is empty, stack $i$ loses one block every second while it remains nonempty. Induction shows that all stacks eventually disappear.

Two lower bounds are immediate:

• Stack $i$ can lose at most one block per second, so $T_i\ge a_i$.
• At the beginning of its final second, stack $i$ has height $1$. It can lose that block only if a neighbour is already empty, so $T_i\ge 1+\min(T_{i-1},T_{i+1})$.

Surprisingly, there is no additional delay: $T_i=\max(a_i,1+\min(T_{i-1},T_{i+1}))$.

We cannot justify this by assuming that a stack shrinks continuously after it first starts: a stack may shrink, pause, and resume. To prove the equality, look at its last pause.