Count the number of integers satisfying two conditions using DP Given two integers $n$ and $m$, how many numbers exist such that all integers have all digits from $0$ to $n-1$, the difference between two adjacent digits is exactly $1$, and the number of digits in the integer is at most $m$?

The integer cannot start with a $0$. All digits from $0$ to $n-1$ must be present.

Example: for $n = 3$ and $m = 6$ there are $18$ such numbers ($210, 2101, 21012, 210121 \ldots$)

I know there is a dynamic programming method to solve this. After looking the solution, I am not able to understand it. Can anybody please give any good solution to me?
Hint: The table consists of cells $T(S,d,\ell)$, which count numbers of length $\ell$ in which the last digit is $d$ and the set of digits is $S$. Unfortunately this solution is exponential in $n$.