Python Practice Problems! Advanced String Functions and Multiple Solution Scenarios - Episode 5

hello and welcome back to the channel this will be episode 5 in our practice problems for python interviews and test and exam prep series and uh today we'll be taking a look at some trickier string operations as well as a scenario where you might be asked to solve multiple solutions of the same problem so if you haven't seen any of the previous videos for this one you don't really need to these are some slightly more complex string operations and then just some scenarios that i've actually had in software interviews so without any further ado let's go ahead and get into it problem given a string return a resultant string where every character inside the string is doubled example pizza turns into p p i i z z z z a a cat turns into c c a a t so this is a fairly straightforward one but if you're not familiar with kind of the concept of for loops or indexing inside a string it can be kind of tricky so that's the only clue that i'll give and if you'd like to attempt this one before we go through it together i recommend go ahead and pause the video now otherwise we'll take the look at the solution together okay so let's go ahead and get started on this one and i'm going to start by defining a function that i will call double string and given an input we are going to make a resultant string that will start out empty and we will go through and for i in range input so we're creating a for loop here that's going to run once per um oh i'm sorry this needs to be length of input so we're creating a for loop that's going to run once per character in the input string so if you think about the prompt what we're supposed to be doing is taking every character in a string and returning it twice so to make sure it runs the right number of times you need something that runs once per character in the input string and what are we going to do in here well we're going to take our result string and add whatever was previously in it to whatever item is currently in that index location i and we're going to do that twice so all we're doing is basically taking one character at a time and adding it twice to a resultant string and then all we need to do is once that's done we just go ahead and print the result and that's really it so let's go ahead and run this function with the two that we were given as examples initially so we can compare them to what's written up top let's do double string with pizza and double string with cat and let's go ahead and run this and see if i did something goofy or if we nailed it okay so if you see right here p p i i z z z z zz a-a-c-a-a-t-t if we want to run it with something other than we other than what we already had you know throw a few states in there ohio new york let's take a look even down to the space what we should see here yep you got a two character space now because that still counts as a character in strings so right away that's the solution to problem 13 um you know if you did it a different way or you have any questions about this solution go ahead and let me know in the comments otherwise we'll move right on so problem 14 given a string return an integer saying how many times abc appears inside the string so for example this string hey hi abc a c a b c high would give you two back because you've got abc there once and abc there once so two individual occurrences and then inside this string you'd have it once obviously if those characters don't appear consecutively inside a string they won't it you'll get a zero back so that's all the hints i'll give if you want to take a crack at this before we look at it together i recommend pausing now otherwise we'll dive into a solution for this so i'm going to call this function i'm going to start just like normal by defining a function and i'm going to call this one check abc that's a good name for it and we'll take an input just like before and uh in this one we're actually gonna be returning a integer not a string so we'll go ahead and make our initial counter and we'll set equal to zero and kind of just like the previous problem we're going to create a for loop for i in range but think about the length of this one because you're not really checking up to the end of your string you're checking up to the end of your string minus two characters because you're not if you don't really care if the last character in a string is in a because it's the last character in a string you're not going to check for a b and a c after you're already past the length you need so for this length we're actually going to do the input minus 2 and that's how many times we want to check if our character plus the two characters after it are a b and c and that'll make sense in this next line of code hopefully so we're gonna say if input at i plus input at i no dictionary input at i plus one plus the input there we go at i plus 2. if all three of those add together are equal to a b c then we're going to do increment the count by one add add plus equals one need a colon there and at the end of this we're just going to print print the count and that is really all we should need to do so see what do you oh it's just because i used a built-in word okay so let's go ahead and check abc here and for the two inputs i'll actually start again by using the two given as the parameters because if you're told certain scenarios and you're told what they should give you that's not a bad place to start and see if your code's working and then you can run through some additional scenarios so let's go ahead and run this and we'll see if we have goofed it up and we get two and then we get one and that's pretty cool so let's go ahead and change this up and say no a b c and i'll throw some spaces in there and then i will say abc abc b c a b c a abc so that should probably give us three counted kind of quickly four a b c a b oh yeah no there's four all right i'm not crazy cool um that's funny actually bca twice i snuck in abc and didn't even notice so this obviously this works pretty good this is an example of how you would sort of check for a sub string inside of another string you can kind of imagine you could set this code up to where if you were given any string and you wanted to check if it was inside of another string you could set up similar checking code even if it was a second input you could set up similar check code to see if it was equal to your second input string rather than abc and you could do a for loop where for however long your input string was you could break this out even longer so this has a lot of applications particularly in things like checking for duplicate entries in a data table or checking for duplicate usernames passwords things like that checking for a substring is a pretty common and useful scenario so i'm sure there's other ways you could have written this if you have another way you prefer that worked for you go ahead and let me know about in the comments otherwise that's problem 14 thanks problem 15 given a string return an integer saying how many times cool appears inside the string however if the string not appears anywhere inside the string or cool does not appear just return not cool so basically we're either returning an integer if cool is in the string telling you how many times cool is in the string unless not is also in the string or cool doesn't appear at all then you're returning text saying not cool so if you'd like to take a crack at this one before we look at it together i suggest pausing the video now otherwise we'll go ahead and check out a solution for this so to start i'm going to create a function just like always and i will call it cool or not and you're going to be given a string and the first thing i'm going to do in the string because it makes sense to me and again there are so many different solutions to this i'm just going through one that works for me but the first thing that makes sense to me is i'm going to check and see if not is in the string because if not is in the string anywhere it actually doesn't matter if cool is or isn't in the string because we're going to return not cool just because the word not is in there because the two scenarios are either cool is not in the string at all which is return not cool anyways or cool is in the string but not is also in there so in either case we return not cool so i'm going to create a true false checker called not check and we'll have this boolean variable and i'm going to check now for i in range and for this one not is three characters so just like the previous example our length is actually going to be the string minus two because we are checking i and then i plus one and then i plus two to check for n o and t and we will do the same if string at i plus string at i plus one plus string at i plus two and if that's equal to not then we're going to do two things we're going to make the not check equal to true because we've found not and then if string is not oh actually there's one more condition we want in here too we want to say and not check is false because we don't want to keep running this if we've already found not we don't want to return okay we found not multiple times like if you had a string that was not not not not cool you wouldn't want to print not cool four times you just want to print it once and move on so we're going to set not the not checker equal to true and we're just going to print out like defined not cool but then we have to think about the other scenario so now if uh if we went through and we didn't find not then we've got to actually so we don't want it to be an else we want it to be at the end of the for loop we want to say if the not check oh boy having trouble writing an underscore so if the not check is still false after that whole for loop runs that means that they went through the whole string and didn't find not in it then this is where we would want to put all of our cool checking so in here i can say for let's say j just so it doesn't get confusing you could reuse i but i don't want to do that so we'll say j in range and now we're checking something that's going to be the length of our string minus um because now we're checking for cool and so we're gonna i'm gonna go ahead and start by copying this guy because it's actually fairly similar so if the string at j plus the string at j plus one plus the string at j plus two get rid of this checker and we need to add in plus the string at j plus three but if all four of those added together is equal to cool then we want our count so that's another variable we should define initially count is initially zero and we want our count to go up by one every time it finds cool inside of the loop and now what we'll say in here after the for loop is done we will say okay if count is if count is equal to zero then we are going to print not cool and then we will say else print and we're just going to print count and actually these should both be inside of the if statement so that we don't run this guy and then run this anyways so let's see that's kind of a lot uh kind of a lot that we just threw in here and i'm pretty good at making silly syntax errors so let's go ahead and check cooler not against a couple strings let's do one that just says cool code man that should give us just a one let's do cool or not and we'll say totally not cool man that's probably just gonna tell us not cool and let's do one that has to count a few times we'll say cool cool cool so we should get one not cool and then three hopefully let's go ahead and run this see how we did one not cool and three hey not bad for a first check um so it this is kind of just what makes sense to me a good way to do it again this problem has at least a dozen different ways you could go about it but the basic check is you want to make sure the knot condition is not in there before you do all the work to check for cool because you there's really no point in just counting up counting up how many times cool is in there and you do all this extra computing if not was in there the whole time and you knew your outcome was going to be not cool it's like trying to check out the wiring for a light switch before you even have like power available to the building and it's like there's no way to know your wiring is correct if the power is completely off so that's a little bit of a tricky one but it's a pretty cool scenario it kind of builds on the previous example where you check for a string inside of another string this one's kind of conditional checking and deciding what to do based on a subset so that's a pretty cool scenario and hopefully you found it useful if you have any questions or you think there's a totally better way to do it just let me know about it in the comments otherwise we'll move on okay problem 16 in the range 1 to 100 if the integer is a multiple of 3 return rocket if the integer is a multiple of 5 return fizz and if the integer is a multiple of 15 return rocket fizz otherwise just return the integer so they're telling you a few things here they're telling you right away you have to run this loop a hundred times they're telling you what to output on multiples of three print out rocket on multiples of five print out fizz multiples of three and five basically fifteen return rocket fizz otherwise just print out the number so they give you a lot of definition but there's really quite a few ways you could go about doing this so if you'd like to take a crack at it i recommend pausing the video now otherwise we'll go ahead and take a look at a solution together okay so i'm gonna just start right away with a for loop because it doesn't really make sense to create a function for this one because there's no parameter game passed in you're sort of told how it needs to run and it's going to run the same time every way there's there's no like given a string or anything like that so we're just going to start with a for loop and say for i in range and it's going to run hundred times uh you know starting at one going to a hundred the cool thing about ranges is it really runs up to that number of times um but it's non-inclusive so i in range 100 is going to run from 1 to 100 or you can think about it 0 to 99 either way we're going to start then by incrementing our counter right in the beginning um actually don't need to do that in a for loop silly uh so what i'm going to do to check is i'm going to check and see if i divide by 3 has no remainder so this is the python function for remainder when you divide an integer so you gotta think this i is going to be um one two three four five six seven eight nine any time it is a multiple of three the remainder is going to be zero so first condition was if that's true and and if i divided by five does not have a remainder of zero so this is saying it is a multiple of three it's not a multiple of five then what are we going to do we're going to print out uh rocket correct yeah and now we're going to copy this condition a few more times because in total there's four things we could return and the basic format is going to be pretty similar for all of them except the last one will just be an else so now let's say if it's not a multiple of three and it is a multiple of five then we are returning fizz um and now you could simplify this to say if it is uh and let's make these l if's um it should be mutually exclusive the way we wrote our conditions but we never want this to run and do two of these in the same time so you could write this as if it is a multiple of 15 or since we already have this framework created you can just do it as it is a multiple of three and a multiple of five and in that scenario we're going to print rocket fizz and otherwise we are just going to print our counter variable and this should be all we need to run let's take a look there is a chance maybe this is going to run 0 to 99 so let's just run it that's the nice thing about code yeah so it looks right there like we stopped at 99. so actually the incrementing of the counter right in the beginning is what you want to do to make sure it actually runs uh one to a hundred rather than zero to ninety nine okay so let's go ahead and just look up at the top and we'll scroll through a little bit of it we see our first text at three and we get rocket which is great because it's a multiple of three and then we get uh at five we get fizz six is rocket rocket fizz threes and fives and then let's go down to our first 15 rocket fizz that looks good we see the numbers in between let's go down to 30 rocket fizz cool so uh we can see that code's working pretty well um this is an actual question i got asked interviewing for a python software position and um you know the this actually goes with the next problem here but um this is just kind of a typical interview question it's not hard they give you a lot of parameters and they just kind of want to understand your thought process so that's one way to do it it worked for me one thing about interview questions in particular and like test and exams where you have a computer in front of you go ahead and run it and see what you get and even if you're on an interview you know as long as you're familiar with the basic steps of troubleshooting and you can talk someone through your process about how you're going to go through and catch stuff usually seeing that you're a problem solver with an intention on figuring out what's going wrong is more important to these people than seeing that you ace the code on your first try um so kind of there how we saw it went 0 to 99 and then quickly fixed it to be 1 to 100 that's a pretty normal thing to have in an interview so that's my solution that's my advice to you guys let me know what you think in the comments and we'll move right along okay problem 17 is the exact same prompt as problem 16 but now you have to do it a different way and again this is actually what i was asked um after creating code and feeling so triumphant about having a solution the interviewer was like okay that was cool um do it another way you know so using a different concept and that's pretty normal if you write code and you're happy that it works but then it's not optimized or it doesn't play nice with someone else's code i mean typically you don't create software in a vacuum you're on a team and so it just it pays to show you can find multiple solutions to solve the same problem so i recommend go ahead and try pausing it and doing it a different way from how you did it previously or how we went through it together previously and take a crack at it and then we'll take a look at another solution together okay so in my mind what we did previously is we just did a for loop for every integer and we used um remainders in division to check if we were on a certain count and what i'm going to do this time is i'm actually going to create a couple counters that you can think of like uh like on an on a um you know like combination lock once you turn something over right like you go zero one two three four five six seven eight nine but it's circular and you turn it over and the next number's nine so i'm gonna create a couple of counters we'll do a three counter and we'll call this other one five counter and essentially we're still gonna need a for loop um and we can say for i in range 100 um or i mean sometimes they'll say even you know they don't want to see the same for loop you could just say while i is less than 100 and you just start by defining i is zero and then walleye is less than 100 we'll increment i just like before we'll and then we'll start doing some checks and we'll say if 3 counter is equal to three and five counter is not equal to five so we're kind of creating those same scenarios as before but now we're doing it right with counter variables and so what we want to do in the situation where we print out rocket we also want the three counter to reset back to one and we want the five counter to just increment okay so because we're not using a for loop and right away you can see this is a lot more code um you don't have to worry about like if you're asked to provide an alternate solution and it sucks like it's a worse solution that's you don't have to worry about that i mean they're trying to see that you can solve something multiple ways they're not trying to test you and see if there was actually a better solution so think about the scenario where now we're saying that it's the five counter we want to print out fizz let's make it an l if just like before and now let's consider the scenario where they're both true and that's when we print out rocket fizz and both counters get reset to one and then the other scenario is going to be none of that's true and that would be a scenario where we increment both these guys and just print out the counter so you print i and you increment both counters and let's check here rather than spend a bunch of time reviewing if i've missed something i'm just gonna look at the solution and see if we got it so the end looks pretty good let's go ahead back up to the start and check again so at three we get rocket five we get fizz six we get rocket 15 is rocket fizz let's go 30 rocket fizz 33 rocket 35 fizz cool so again i like the first way we did rocket fizz um i liked the for loop it was way less code you can tell just click in between the two we did this in seven lines of code and we did this in 20 i mean this is not optimized um but you never know like for the application that you need code written in there are better and worse solutions and this is a pretty good example of you're going to be asked to solve the same problem two different ways and a lot of software guys a lot of tech guys they get very stubborn in like well i know how to solve the problem but a hiring manager or an evaluator might be trying to see that you can think creatively of multiple ways of solving something so hopefully you found that useful you know if you solved it a totally different way if uh you have any questions about anything you saw here just let me know in the comments um otherwise if you found this or any content on the channel helpful go ahead and leave a like and a subscribe i really appreciate that helps the channel out a lot and as always good luck good luck with your code and thanks for watching thanks bye