codejam-2021 | Python 3 solutions to Google Code Jam 2021 problems | Learning library

This rather terse code implements a dynamic programming algorithm that has been optimised to reduce the space usage from O(n) to O(1). First I'll describe an equivalent DP and how it can be computed, and then show how the many separate cases it needs to handle can be viewed as instances of a smaller number of more general "case templates", which kamyu's code exploits for concision.

The question frames the problem as finding a minimum-cost way to replace each "?" character with a J or C, but we can alternatively frame it as finding a minimum-cost way to replace all characters of the input string S, provided that we never change a J to a C or vice versa -- that is, we can imagine ourselves "replacing" an existing J with another J, or an existing C with another C. In fact, that "provided" clause can even be removed: We can get what we want by permitting all possible replacements, including a J to be changed to a C or vice versa, but prevent these undesirable changes from actually appearing in any optimal solution by penalising them with enormous costs.

Let's define the function f(m, a) to be the minimum cost of a solution to the subproblem consisting of the first m+1 characters of the input string S, under the assumption that we replace the final (i.e., (m+1)-th) character with a, which must be either "J" or "C". (Why m+1 and not m? That just makes code with 0-based array indices simpler.) Replacements that leave this character unchanged (J->J or C->C) are allowed, as are replacements of "?" characters to either J or C. The way we will prevent changing an existing J to a C or vice versa is to assign infinite cost to such a replacement -- this will result in this choice never being taken, since another lower-cost choice is always available.

We can compute values of f(m, a) as follows. I'll use Python-like pseudocode that writes to a DP matrix dp[][].

The highest-priority rule is that we should always assign infinite cost to any attempt to change a hardwired letter to the other letter. Other than that, the cost of assigning a particular letter depends on the preceding letter -- unless there is no preceding letter because we are at the very start, in which case we know that the cost is zero: