First of all, we should note that ~\bmod 10~ is just taking the unit digit (with the special case described for ~0~). The unit digits of the powers of a number follows a pattern. Additionally, only the unit digit of the base of the power is relevant, meaning ~26^N \bmod 10~ is equivalent to ~6^N \bmod 10~. Either storing the number cycles or modding the power by 4 (the LCM of the cycle lengths of the digits) is sufficient to pass.

Time complexity: ~\mathcal{O}(L_T)~ where ~L_T~ is the total number of letters in both names.