= Solution
\b[Yes: starting from zero, $n$ increments of the <decimal counter> cost $\Theta(n)$ <digit> changes], uniformly in the number $k$ of <digits>. An increment changes the trailing <digits> equal to $9$ into $0$ and increases the next <digit>. Number <digits> from $0$ at the least significant end. <Digit> $i$ changes once every $10^i$ increments, so the exact number of changes is
$$
C_{10}(n,k)=\sum_{i=0}^{k-1}\left\lfloor\frac{n}{10^i}\right\rfloor
\leq n\sum_{i=0}^{\infty}10^{-i}=\frac{10n}{9}.
$$
Here the <floor function> counts carries reaching each position, and the last equality sums a <geometric series>. Since <digit> $0$ changes on every increment, $C_{10}(n,k)\geq n$ for $k\geq1$. The loop tests and carry handling add only a constant amount per changed <digit> and per operation, giving the same <time complexity>.
An <accounting method> explanation is also possible. Charge two credits for each change of <digit> $0$: one pays for that change and one joins a common bank. For $n$ increments the higher <digits> change at most $n/9$ times, so their costs can be covered from that bank; the displayed prefix bound ensures that it never goes negative.
The formula allows wraparound modulo $10^k$; alternatively, restrict to increments before overflow. Initializing the <array> itself costs $\Theta(k)$ if that is counted separately. If the initial <digits> are arbitrary, the number of changes at position $i$ is at most $\lceil n/10^i\rceil$, giving $O(n+k)$ in <Big O notation>. Starting with all <digits> equal to $9$ makes the first increment cost $k$, so the initial-state convention matters.
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