Digit DP
Most of the time, you will have lower bound and upper bound.
The good solution might be
f(x) => answer for x
ans = f(upper) - f(lower - 1)
How to think of digit dp?
define the function first, you should have at least 2 params
backtrack(lens, tight)
# lens equal to current digit
# tight means if we need to take care of the bound limitation