You're going to multiply pi with pMiss.

Okay, let's do insertion first. It does an insertion in this list. That insertion takes Θ(k).

So I don't want my_sort() to return a copy of the list. I want, after I call my_sort() on a , if I then refer to a

Now, selection is the same.

And there's tradeoffs to doing it this way.

What does that give us for a running time for this entire algorithm?

So let's run this sample code and see what happens.

Now we're reassigning what p points to to the original value of p, which is this, concatenated with the list 1.

Is that constant time, log time, or linear time? Same thing for an unordered list.

Ah, and we see our link id 24.

And it's monitoring their reliability.

The actual list is secure at Langley.

The part of the genome that we can read is the list of ingredients.

Inserting a new element into the list and it has to stay ordered now because this is an ordered list that involves running to the list or possibly jumping around in the list.

I'm hoping if you've been paying attention that that was pretty easy to see that that's linear.

This is going to change the elements in p.

This is similar to the previous question where we want to go through the list one entry at a time.

Simple enough.

We can print just 1 of the elements.

We have a for statement where we'll introduce the name e as the variable name.

In that example, I kept track of both of the best and the second best, and then it was running through the list.

let's say I do print(a), if I do print(a) then that will literally return a sorted version of a .

Start set is over here, goal set is over here.

Run Execute the program

Again, the upper left corner is the Program display box

But I'm going to define my vector.

Do It

Run as different user

Run On

Execution

Run

Execute

Exec

Go.

Execute Executes the current line in the console

In fact, I think its probably not constant time, but you can create a version where each of these steps is constant time so that this entire routine runs in Î n because all we're doing is we're touching each of the values in the list to construct a new list

The top k algorithm that we have so far.

We do get the value 12.

Alright. So now we consider the unordered case.

It just means a list of two numbers.

In this case, all of these operations run in Î (1) constant time, given that the list is sorted.

Let me ask you a trick question.

Execute Command

running

Binary