jkrt's AND

First, let us characterize all positive integers satisfying the condition.

Every one-digit positive integer $1,2,\ldots,9$ satisfies the condition. Indeed, $x$ and $x-1$ are different one-digit numbers, or $x-1=0$ when $x=1$, so their LCS length is $0$.

Now suppose $x$ has at least two digits. If the last digit of $x$ is not $0$, then $x-1$ is obtained by decreasing only the last digit of $x$ by $1$. Therefore, the prefix before the last digit remains unchanged, and this prefix is nonempty. Hence $x$ and $x-1$ have a common digit, so their LCS length cannot be $0$.

Thus, any valid $x$ with at least two digits must end with $0$. Remove as many trailing zeros as possible and write

$$x = y \cdot 10^t \quad (t \ge 1),$$

where $y$ does not end with $0$.

If $y$ has at least two digits, then $y-1$ is obtained by decreasing only the last digit of $y$ by $1$, so a nonempty prefix remains unchanged. Therefore, $x$ and $x-1$ still have a common digit. Hence $y$ must be a one-digit number.

So every valid number with at least two digits must have the form

$$x = d \cdot 10^t.$$

Here, $1 \le d \le 9$ and $t \ge 1$.

If $d=9$, then the decimal representation of $x$ contains digit $9$, and the suffix of $x-1$ contains digit $9$, so they have a common digit. Therefore, $d=9$ is impossible. Conversely, if $1 \le d \le 8$, the digits of $x$ are only $d$ and $0$, while the digits of $x-1$ are only $d-1$ and $9$. These sets are disjoint, so the condition is satisfied.

Therefore, the valid numbers are listed as follows.

$$1,2,\ldots,9,\quad 10,20,\ldots,80,\quad 100,200,\ldots,800,\quad 1000,2000,\ldots$$

If $N \le 9$, the answer is simply $N$.

If $N \ge 10$, then

$$A_N = d \cdot 10^e,$$

where

$$d = ((N-10) \bmod 8) + 1$$

and

$$e = \left\lfloor \frac{N-10}{8} \right\rfloor + 1 = \left\lfloor \frac{N-2}{8} \right\rfloor.$$

Since $N$ can be very large, we process it as a decimal string. To compute $d$, it is enough to compute $N \bmod 8$. To compute $e$, subtract $2$ from the decimal string and divide it by $8$, keeping only the quotient modulo $998244352$.

Finally, by Fermat's little theorem,

$$10^{998244352} \equiv 1 \pmod {998244353}.$$

Therefore, when computing $10^e \pmod {998244353}$, it is enough to know $e$ modulo $998244352$.

The time complexity is $O(|N|+\log 998244353)$, and the memory usage is $O(|N|)$.