Let \(f_{AB}(i)\) be the number of Bs minus the number of As among the first \(i\) characters of the string, and let’s define \(M_A = max(f_{AB}(0), f_{AB}(1), \ldots, f_{AB}(3N))\). Let’s define \(f_{BC}\) and \(f_{CA}\) similarly, and \(M_B = max(f_{BC}(0), f_{BC}(1), \ldots, f_{BC}(3N))\), \(M_C = max(f_{CA}(0), f_{CA}(1), \ldots, f_{CA}(3N))\).

We will prove that it’s possible to split the string \(T\) into such \(N\) subsequences if and only if \(M_A+M_B+M_C\le N\).

Necessity: If there exists such splitting, then \(M_A+M_B+M_C\le N\).

Suppose that we have managed to split our string into \(A_1\) subsequences ABC, \(B_1\) subsequences BCA, \(C_1\) subsequences CAB. Then the balance \(M_A\) can’t be worse than \(B_1\), as in subsequences ABC and CAB A goes before B. Similarly, the balance \(M_B\) can’t be worse than \(C_1\), and balance \(M_C\) can’t be worse than \(C_1\), so \(M_A+M_B+M_C\le A_1 + B_1 + C_1 = N\).

Sufficiency: If \(M_A+M_B+M_C\le N\), we can split letters into triples (\(i\)-th A, \((i+M_A)\)-thB, \((i+M_A+M_B)\)-th C) (here \(i+N\)-th letter \(X\) denotes \(i\)-th letter \(X\), meaning that we enumerate cyclically).

Let’s consider string \(S+S+S\). Note that in it for any \(i\le 2N\), \(i\)-th A goes before \(i+M_A\)-th B, \(i\)-th B goes before \(i+M_B\)-th C, \(i\)-th C goes before \(i+M_C\)-th A (from the balance logic). This means that \(i\)-th A goes before the \(i+M_A\)-th B, which goes before the \(i+M_A+M_B\)-th C, which goes before the \(i+M_A+M_B+M_C\)-th A, which coincides or goes before to \(i+N\)-th A. This clearly means that \(i\)-th A, \((i+M_A)\)-thB, \((i+M_A+M_B)\)-th C go in the cyclic order in the string, so the matching is valid.

With this observation, writing \(O(N^6)\) dp is trivial: just keep track of how many letters A, B, C we have already taken, and the max values of \(f_{AB}(i), f_{BC}(i), f_{CA}(i)\) so far. At first sight, it’s \(O(N^7)\), but the total number of characters after processing \(i\)-th one has to be \(i\), so it’s still \(O(N^6)\) with an extremely small constant (the solution would work for \(N\le 40\) as well).

Bonus: What if the problem is about splitting string of \(N\) As, \(N\) Bs, \(N\) Cs, \(N\) Ds into subsequences ABCD, BCDA, CDAB, DABC?