If n is the number of keys in the dictionary, then the whole procedure would not be O(n log n), it'd be O(m log n) where m is the number of letters. Given that

If n is the number of keys in the dictionary, then the whole procedure would not be O(n log n), it'd be O(m log n) where m is the number of letters. Given that n >> m, I would think your original statement is correct.