 Hello again, class. In my previous video for part two of problem one in my final exam review, I showed you how to solve the sequence of tokens returned by continuously calling getToken using the fully fleshed out programmatic approach, which is very important for you to know. It's important to understand. However, you might have noticed that it also takes a long time to run through. And when you're completing your final exam, you have very precious little time for each problem. So I'm going to show you a faster approach that you can consider adopting to help save you some time. Instead of creating a full table and stepping through character by character, I'm simply going to write out the string that we're given. Now, I'm going to starting at the beginning of the string, I'm going to step through each regular expression and see the longest substring that each regular expression will match of our full string. So letter won't match, neither will capital letter, but digit will match the first zero. So we have digit here matching a string of length one. Alpha has letter star, which can match the empty string, followed by a digit. So that can match the first zero, followed by zero or more capital letters. So that can match the empty string, and we see that alpha does in fact match a substring of this of length one. Next up we have row. Digit star can cover the first two zeros in the string, followed by zero or more capital letters, which won't match the lowercase a, but the capital letters star can still match the empty string. So we see that row has a match of length two. For phi, phi requires two digits, followed by zero or more lowercase letters. So the first digit will match zero, second digit will match zero, and for zero or more lowercase letters, we can match the lowercase a. And we see that phi has a match of length three. Finally for omega, letter star can match the empty string, followed by a letter or zero or more digits. So for zero or more digits, this will cover the first two zeros. And we have a match of length two. We'll notice that phi was the longest match of our regular expressions of length three. So we are going to cover the first three characters in this string and match them to phi. Then we'll start over beginning with the capital D. The lowercase letter regular expression won't match it. Capital letter will match the capital D. Digit won't match any. And alpha can match zero letters, followed by a digit or a capital letter. So the capital letter here can match the capital D, followed by zero or more capital letters. So we are at the end of the regular expression, but we only matched a substring of length one. For row, row will match zero or more digits, followed by zero or more capital letters. So it'll match the capital D, and we'll have a match of length one. Phi requires a digit as the first symbol in the string, so it will not match. And omega will take zero or more capital letters, so that can match the capital D, followed by a letter or zero or more digits. So this letter will match the lowercase a. We're at the end of the regular expression, and we have a match of length two. Omega was our longest match here. So we will take the next portion of the string, the substring of length two, and match it to omega. Finally we'll step through one more time. We see that letter will not match, capital letter will not match, but digit will match the three, so we have a match of length one. And then alpha will match zero or more letters. So if this is the empty string, followed by a digit or a capital letter, so this digit will match three. Followed by zero or more capital letters, so this will match the capital D. We'll see that alpha fully matches the rest of the substring, with a length of two. Now because we've completed the string, we don't need to go through and check each of the other regular expressions, because we have a rule that states that if we have any ties that match the same length, ties are broken in favor of tokens that appear first in the list. So alpha was the first regular expression to match the rest of the string, and we can just take alpha. And this gives us our final answer of phi omega alpha. This is clearly a much faster approach to what I showed you last time, however there is a drawback to it. And that's if, if you make a mistake, Dr. Dupay won't be able to tell where you made the mistake, and so you lose the ability to gain partial credit on the problem. However doing this does save you a ton of time, which you might want when completing the rest of the exam. I hope this helps, and I wish you all the best of luck.