— And then keep list sorted
And check at the middle of the list if the first char matches, then go to 1 / 4 or 3 / 4 based on higher / lower, then repeat for the next level, and keep going
— You could make a binary search tree.
— But it's not space efficient.
— I was thinking a linear list
— But looking up list.length / 2 first, then the next half
— You could try to implement your own hash map.
— Would be faster than binary search.
— Well, don't want to mess with hashing
— Want to do it without needing that
— I decided we don't need to do index assignment
— We can store the strings as strings, but also keep the unique characters recorded