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Describe an in-place version of the quick-select algorithm in pseudo-code....

Question

Describe an in-place version of the quick-select algorithm in pseudo-code.

Describe an in-place version of the quick-select algorithm in pseudo-code.



Answers

Describe an algorithm that takes as input a list of $n$ integers and produces as output the largest difference obtained by subtracting an integer in the list from the one following it.

Okay, So here were describing an algorithm that will find our repeating introducing a list. Right? So let's just go ahead and call this procedure repeated something that explains what does and this procedure will take in a list of integers. So up to some n number of integers, I was gonna shortened integers two ends here. And of course, this will only work if there's at least one element in the list. Right? So we need and to be greater than or equal to one. And so this w slash just meets with. Additionally, it is a non decreasing list. So a one must be less than or equal to a two, which must be less than or equal to a three. How much must be less than equal to a four and so on sees the stipulations we have. And since we are treating a list of these numbers were going to initialize an empty set that we can add Thio. Additionally, we're gonna create an N plus one element to our list. Right? So all set a n plus one equal to a n this last moment in the list. Um, we're gonna subtract, attract one from it. So it's less than that. So you don't be confused here. And plus one is our sub script. Um, but then the ai n just the end is the sub script, and we're subtracting it. And so this By creating this last element, we do that due to the comparisons we're going to make. So, while investigating any element, we're going to compare it to both the previous and next element in the list. All right, so they have it listed things here. The problem is, once we get to the last element in the list, their ends up not being anything for us to compare it to. Okay, So where initialized We're just sort of creating something there that will Paris our cellist, um, in an appropriate way. Okay, so now that we have lived in that initialized these elements, we're going to run through the list, right? So for I that's gonna actually started thes second index right And does that because again, we're comparing to the previous previous items in the less sort of started it a two, and it's gonna run on the way to end. And if if the element we're currently investigating is equal to the previous one. It says. That's just checking to see if we have. If we're currently have a two, we're just seeing Haze. That is the thing before it also it too. Okay, that's that's not quite enough. Because if we had three twos, if all we did was check the previous one, we'll be adding to to our list of repeated elements twice, right? Both at the second time it occurs, and the third time it occurs so that make sure you only out it to our list once. We also want to check whether or not the next item is larger. Okay, so we represent that by saying air is less than a I plus one, right? This is it. This is why we had to do this initialization here in the second step. And if those both hold, then we want to add the element in question to our list. Right? So we'll take the list and said equal to itself Whatever is currently in there, plus aye aye. And then finally, we want to return on our list. We wantto see all the things that are repeated in our list that we at submitted in twos algorithm. So we want to return s. And that will complete

Okay, so for this problem, we wanna take a list of their input and just figure out how many of the introducing the list are Negatives will call this algorithm, count an egg for count. Negative. And our input is this list, right? So it's gonna be a list of numbers are in this case, it's into yours, and our list is going to be, uh, and elements long. Okay, These are integers, and it's also important to note that there should be at least ah, one element in the list. Okay, so not an empty list. What is the point here? Is that, uh, we're going to be counting something. And so to do that well, you want a counters will set K is our counter. I want to say equal to zero Rex. There might not be any negative numbers in the list. All right, then, if we let I be our index for the latest still safe for I mean, this is equal to one. So to start, and then it'll go up to end so equal toe I've up to and well say, if the element A I is less than zero. Well, then we want to add one to our counter. Right? So that will, well set k equal notice. Trying to add once will take whatever its current value is and add one to it. At the end of this, we will return kegs. That was what was counting are negative numbers. All right, So if we set a counter equal to zero here to start and then reiterate to the list of for every element in the list, if that element is less than zero, add one to the counter at the end of going to the entire list, return the counter.

Okay, so here we again are having a list as input, and we're trying to find the largest even interview. So well, color our algorithm, uh, large even. And the input here is going to be a list, right? So we'll have our list of integers. And as in the previous examples, we will you read them with is a little too and then So these are our integers, and this shouldn't be in empty list. So we also want to specify that and should be at least equal to one. Okay, so they're gonna have a few counters, right? So we'll send k equal to zero. And this is this is just going to keep track of the index or the location of our largest even number. And then we'll also have. I'll say I'll say J Okay, this is gonna be the value, are the largest, even integer. That's when you start the Mets zero. So if it goes through the whole list, it should be returning a zero, right? Since we want to set these storm off at zero, okay, and they will say, Well, four it's reading out that words afore I in our arranging indexes that's going to 0 to 1. All right there. So for one to end for eye in the range of one two are in next end. We need to check a few things, Andre. So first of all, Ah, I I needs to. Even so if if it's even, and you could do that by saying you know there's no remainder when it's divisible by two divided by two of us and only doesn't need to be even it should also be larger than our J value, right? So and I should be larger than whatever current largest, even number is. Well, they will set, said J. Equal to a ice that's keeping track of what our current largest value is. We're also going to set okay, equal to ay, right, says the index that were at the location of this largest number. And at the end of going through all of these, well, the point was to return the location rights we want to return Okay, which is the location of it. But we're using

All right. So here we are wanting to insert into during to list, uh, other into juice. We're just gonna call this one insert, but you can call it really whatever you like. And the input is going to be a list. It's just gonna be a list of into juice. End of them. I was gonna abbreviate into juices. We do require that there's at least one element in the list and also that they're in increasing order, writes a one would need to be less than equal to two, which needs to be less than equal to a three and so on when they close this bracket too soon mix. Of course, we also have another item that this would take enough the interview that we're searching for. Right? So ex also into jail. But this is what we're going to be inserting into our list. So initially we're going to sort of add a last element. Okay, we're gonna do that. So there's always an integer larger uh, then X in the list. This will help us be able to insert accident sponte, assuming that it is larger than any element in the list. So what? It is to find a and plus one as being larger than X, and we're gonna replace this later by its exact value. So don't worry about adding to the list unnecessarily. And so the list of integers are already in increasing order. So we're gonna start by Iterating threw them right in the 1st 2nd then the third we want an index to keep track of. That sort is gonna set I index equal the ones we're starting at the first moment in the list. And so, while ex thea animatronics search is greater than any given index, I'm gonna change I to the next one, right? So I'm just gonna set it equal to itself, plus one more and so X is larger than a one. Then we go to the next, next to and in excess larger than a two. We would go to the next next three and so on, and at some point that in next will stop. It will be smaller than something, right? And once we hit that point, we will shift all the values larger the next, the next position, right, And this will give us a place to insert X. So for kay being equal to zero to end minus I. So now just shifting. All right, so whatever the A to the n minus K plus one it is. It's gonna be moved over one too, eh? On minus K. Okay, So now we can put X into its position at a I all right, in this shifting right, the shifting that occurred here will replace this initial element. So we don't have to worry about having added something that doesn't belong, Okay? And so now that everything is added, the last thing to do this last step I've run out of room for all right. Adoption society is, of course, to return our list rights. We just want to return the list at wanting to etcetera only down to a n plus one, right? It's plus one because we've added to us. There's one more element than there was before.


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