Note that the original editorial was in Lithuanian and may have translation errors.

The problem is solved by generating all variants and applying dynamic programming, to avoid having to recompute everything.

A dynamic table ~\text{dp}[i_1, r, i_2, i_3, i_4, i_5]~ is created; It shows the number of ways in which a phrase can be completed if:

• already completed up to and including the ~i_1~th letter;
• the ~i_1~th letter is ~r~;

Counting backwards from the ~i_1~th letter (we have no idea what is happening at the beginning of the word only what happens at the end, i.e. at the point where we need to fill in from) is

• ~i_2~ identical successive consonants
• ~i_3~ consecutive consonants
• ~i_4~ identical successive vowels
• ~i_5~ consecutive vowels

For example, the initial phrase y*af:

We filled it up to aayb in a recursive way, i.e.

• ~i_1 = 4~,
• ~r =~ b,
• ~i_2 = 1~ (ending in a single consonant b)
• ~i_3 = 2~ (ending in two consonants yb);
• ~i_4 = 0~ (no vowels at the end);
• ~i_5 = 0~.

Instead of all the *'s, we try to substitute all the possible letters from a to z. For each letter, we check whether it can be done (i.e. whether it does not exceed the set norms), and fill in the table. If we can take a look from the table, we take it from there. If not, it is calculated recursively and stored in the table.