शैडोइंग अभ्यास: 2. Data Structures and Dynamic Arrays - वीडियो के साथ अंग्रेजी बोलना सीखें

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Good morning.

Welcome to 006

Introduction to Algorithms lecture two.

I am Eric DeMaine and I love algorithms.

Are you with me?

Yeah.

Today, we're not doing algorithms.

No, we're doing data structures.

It's OK.

There's lots of algorithms in each data structure.

It's like multiple algorithms for free.

We're going to talk about sequences and sets and linked lists and dynamic arrays.

Fairly simple data structures today.

This is the beginning of several data structures we'll be talking about in the next few lectures.

But before we actually start with one, let me tell you slash remind you of the idea, the difference between an interface,

which you might call an API if you're a programmer, or an ADT if you're an ancient algorithms person like me, versus a data structure.

These are useful distinctions.

The idea is that an interface says what you want to do.

A data structure says how you do it.

So you might call this a specification.

And in the context of data structures, we're trying to store some data.

So the interface will specify what data you can store,

whereas the data structure will give you an actual representation and tell you how to store it.

This is pretty boring.

Just storing data is really easy you just throw it in a file or something.

What makes it interesting is having operations on that data.

So in the interface, you specify what the operations do,

what operations are supported, and in some sense, what they mean.

And the data structure actually gives you algorithms.

This is where the algorithms come in for how to support those operations.

All right.

So in this class, we're going to focus on two main interfaces.

and various special cases of them.

So the idea is to separate what you want to do versus how to do it, because you can think of this as a problem statement.

So last class, Jason talked about problems and defined what a problem was versus algorithmic solutions to the problem.

And this is the analogous notion for data structures, where we want to maintain some data according to various operations.

So the same problem can be solved by many different data structures.

And we're going to see that.

And different data structures are going to have different advantages.

They might support some operations faster than others.

And depending on what you actually use those data structures for, you choose the right data structure.

But you can maintain the same interface.

We're going to think about two data structures.

One is called a set and one is called a sequence.

These are highly loaded terms.

Set means something to a mathematician.

It means something else to a Python programmer.

Sequence, similarly.

I guess there's not a Python sequence data type built in.

The idea is we want to store n things.

The things will be fairly arbitrary.

Think of them as integers or strings.

And on the one hand, we care about their values.

And maybe we want to maintain them in sorted order and be able to search for a given value, which we'll call a key.

And on the other hand, we care about representing a particular sequence that we care about.

Maybe we want to represent the numbers 5, 2, 9, 7 in that order and store that in Python.

You could store that in a list, for example.

And it will keep track of that order.

And this is the first item, the second item, the last item.

Today we're going to be focusing on this sequence data structure, although at the end I'll I'll mention the interface for sets,

but we're going to be actually solving sequences today.

And in the next several lectures, we'll be bouncing back and forth between these.

They're closely related, pretty abstract at the moment.

On the other hand, we're going to have two main

Let's call them data structure tools or approaches.

One is arrays, and the other is pointers,

pointer -based or linked data structures.

You may have seen these.

They're used a lot in programming, of course.

But we're going to see both of these today.

I'll come back to this sort of highlight in a moment.

Let's jump into the sequence interface, which I conveniently have part of here.

There's a few different levels of sequences that we might care about.

I'm going to start with the static sequence interface.

This is where the number of items doesn't change, though the actual items might.

So here we have n items.

I'm going to label them x0 to xn minus 1, as in Python.

So the number of items is n.

And the operations I want to support are build, length, iteration, get, and set.

So what do these do?

Build is how you get started.

To build a data structure in this interface, you call build of x.

Exactly how you specify x isn't too important, but the idea is I give you some items in some order.

In Python, this would be an iterable.

I'm going to want to also know its length.

And I want to make a new data structure of size n, a new static sequence of size n that has those items in that order.

So that's how you build one of these.

Because somehow we have to specify n to this data structure, because n is not going to be allowed to change.

I'm going to give you a length method.

These methods are the object -oriented way of thinking of operations that your interface supports.

So length will just return this fixed value n.

Iter sequence, this is the sense in which we want to maintain the order.

I want to be able to output x0 through xn minus 1 in the sequence order,

in that specified order that they were built in or that it was changed to.

So this is going to iterate through all the items.

So it's going to take at least linear time to output that.

But more interesting is we can dynamically access anywhere in the middle of the sequence.

We can get the ith item XI, given the value I, and we can change XI to a given new item.

So that's called get at and set at.

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Pretty straightforward.

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This should remind you very closely of something, a data structure.

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So this is an interface.

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This is something I might want to solve.

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But what is the obvious data structure that solves this problem?

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Yeah?

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A list.

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In Python, it's called a list.

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I prefer to call it an array, but to each their own.

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We're going to use lists.

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Lists could mean many things.

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But the solution to this interface problem,

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the natural solution, is what I'll call a static array.

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Jason mentioned these in lecture one.

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It's a little tricky because there are no static arrays in Python.

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There are only dynamic arrays, which is something we will get to.

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But I want to talk about what is a static array really.

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And this relates to our notion of our model of computation, which Jason also talked about, which we call the word RAM.

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So the idea in word RAM is that your memory is an array of W This is a bit circular.

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I'm going to define an array in terms of the word ram, which is defined in terms of arrays.

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But I think you know the idea.

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So we have a big memory, which goes off to infinity maybe.

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It's divided into words.

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Each word here is w bits long.

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This is word 0, word 1, word 2.

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And you can access this array randomly, random access memory.

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So I can give you the number 5 and get 0, 1, 2, 3, 4, 5, the fifth word in this RAM.

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That's how actual memories work.

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You can access any of them equally quickly.

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Okay, so that's memory.

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And so what we want to do is when we say an array,

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we want this to be a consecutive chunk of memory.

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Let me get a color.

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here.

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So let's say I have an array of size 4, and it lives here.

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Jason can't spell, but I can't count, so I think that's 4.

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So the array starts here, and it ends over here.

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It's of size 4.

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And it's consecutive, which means if I want to access the array at index i,

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then this is the same thing as accessing my memory array at position wherever the array starts,

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which I'll call the address of the array.

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In Python, this is ID of array plus i.

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This is just simple offset arithmetic.

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If I want to know the zeroth item of the array, it's right here where it starts.

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The first item is one after that.

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The second item is one after that.

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So as long as I store my array consecutively in memory, I can access the array in constant time.

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I can do get at

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and set at as quickly as I can randomly access the memory and get a value or set a value, which we're assuming is constant time.

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So array access is constant time.

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OK, this is what allows a static array to actually solve this problem in constant time

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per get at and set at operation.

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So this may seem simple, but we're really going to need this model

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and really rely on this model increasingly as we get to more interesting data structures.

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This is the first time we're actually needing it.

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Let's see.

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Length is also constant time.

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We're just going to store that number n along with its address.

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And build is going to take linear time.

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Iteration will take linear time.

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OK, pretty straightforward.

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I guess one thing here when defining build, I need to introduce a little bit more of our model of computation,

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which which is, how do you create an array in the beginning?

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I claim I could do it in linear time, and that's just part of the model.

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So this is called the memory allocation model.

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There are a few possible choices here,

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but the cleanest one is just to assume that you can allocate an array of size n in theta n time.

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So it takes linear time to make an array of size n.

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You could imagine this being constant.

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It doesn't really matter much, but it does take work.

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And in particular, if you just allocate some chunk of memory, you have no idea whether it's initialized.

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So initializing that array to zeros will cost linear time.

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It won't really matter, constant versus linear, but a nice side effect of this model is that space, if you're just allocating arrays,

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the amount of space you use is at most the amount of time you use, or I guess big O of that.

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So that's a nice feature.

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It's pretty weird if you imagine,

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it's unrealistic to imagine you can allocate an array that's infinite size and then just use a few items out of it.

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That won't be, give you a good data structure.

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So, we'll assume it costs to allocate memory.

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And okay, great.

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We solved the static sequence problem.

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Very simple, kind of boring.

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These are optimal running times.

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Now, let's make it interesting.

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Make sure I didn't miss anything.

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And talk about, oh, there is one thing I want to talk about in the word RAM.

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A side effect of this assumption that array access should take constant time

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and that accessing these positions in my memory should take constant time is that we need to assume

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w is at least log n. or so.

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W, remember, is the machine word size.

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In real computers, this is currently 64 or 256 and some bizarre instructions.

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But we don't usually think of the machine as getting bigger over time.

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But you should think of the machine as getting bigger over time.

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This is a statement that says the word size has to grow with n.

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It might grow faster than log n, but it has to grow at least as fast as log n.

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Why do I say that?

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Because if I have n things that I'm dealing with, n here is the problem size, maybe it's the array I'm trying to store, whatever.

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If I'm having to deal with n things in my memory, at the very least, I need to be able to address them.

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I should be able to say, give me the ith one and represent that number i in a word.

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Otherwise, because the machine is designed to only work with w bit words in constant time,

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if I want to be able to access the ith word in constant time,

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I need a word size that's at least log n just to address the n things in my input.

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So this is a totally reasonable assumption.

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It may seem weird because you think of a real machine as having constant size, but a real machine has constant size RAM also.

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My machine has 24 gigs of RAM or whatever.

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That laptop has eight, but you don't think of that as changing over time.

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But of course, if you wanted to process a larger input, you would buy more RAM.

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So eventually, when our Ns get really, really big, we're going to have to increase W just so we can address that RAM.

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That's the intuition here.

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But this is a way to bridge reality, which are fixed machines, with theory.

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In algorithms, we care about scalability for very large n.

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We want to know what that growth function is and ignore the lead constant factor.

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That's what asymptotic notation is all about.

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And for that, we need a notion of word size also changing in this asymptotic way.

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All right.

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That will be more important next week when we talk about hashing and why hashing is a reasonable thing to do.

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But let's move on to dynamic sequences, which is where things get interesting.

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So I have the update here.

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So we start with static sequences.

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So all of these operations are still something we want to support in the dynamic sequence, but we add two dynamic operations,

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operations, somewhat controversial operations, so exciting.

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I want to be able to insert in the middle of my sequence,

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and I want to be able to delete from the middle of my sequence.

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So here's my sequence, which I'm going to think of in a picture.

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I'm going to draw it as an array, but it's stored however it's stored, we don't know.

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This is an interface, not an implementation.

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So we have x0, x1, x2, x3.

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And let's say I insert at position 2.

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So position 2 is here.

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So I come in with my new x.

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And I would like x to be the new x2, but I don't want to lose any information.

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If I did set at 2, then I would erase this and replace it with x.

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But I wanted to insert at, which means all of these guys conceptually are going to shift over by 1 in terms of their indices.

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So then I would get this picture that's one bigger.

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And now I've got the new x.

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I've got what was the old x2, which I hesitate to call x2 because that's its old name, not its new name.

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I'm going to draw arrows to say these guys get copied over.

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These ones are definitely unchanged.

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Our new x2, which I'll call x2 prime, is x.

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And this is x3 prime, x4 prime, so on.

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I want to be careful here.

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And of course, the new n prime is n plus 1.

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I want to be careful about the labeling, because the key, what makes insert at interesting is that later when I call get at,

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it's with the new indexing.

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So previously, if I did get at at 2, I would get this value.

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And afterwards, if I did get at at 2, I get the new value.

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If I did get at it 3 down here, I would get the value that used to be x2.

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That's maybe hard to track.

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But this is a conceptually very useful thing to do, especially when you're inserting or deleting at the ends.

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So we're going to define,

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in particular, insert and delete first and last.

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So these are sometimes given.

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If you have an insert, it has an x.

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If you do a delete, it has no argument.

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So this means insert at the beginning of the array, which would be like adding it here.

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And insert last means adding it on here.

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So insert last doesn't change the indices of any of the old items.

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That's a nice feature of insert last.

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Insert first changes all of them.

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They all get incremented by one.

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And we're also interested in the similar things here.

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We could do get first or last or set first or last,

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which are the obvious special cases of get at and set at.

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Now, these special cases are particularly interesting in an algorithms context.

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If you were a mathematician, you would say, well, why do I even bother this?

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This is just shorthand for a particular call to get or set.

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But what makes it interesting from a data structure's perspective is that we care about algorithms for supporting these operations.

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And maybe the algorithm for supporting get first or set first

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or in particular insert first

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or insert last might be more efficient maybe we can solve this problem better than we can solve insert at.

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So while ideally we could solve the entire dynamic sequence interface constant time per operation, that's not actually possible.

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You can prove that.

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But special cases of it where we're just inserting and deleting from the end say, we can do that.

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So that's why it's interesting to introduce special cases that we care about.

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Cool.

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That's the definition of the dynamic sequence interface.

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Now, we're going to actually solve it.

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So our first data structure for this is called linked lists.

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We've taken probably, you've probably seen linked lists before at some point.

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But the main new part here is we're going to actually analyze them

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and see how efficiently they to implement all these operations we might care about.

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So first review, what is a linked list?

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We store our items in a bunch of nodes.

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Each node has an item in it and a next field.

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So you could think of these as class objects with two class variables, the item and the next pointer.

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And we assemble those into this kind of structure where we store in the item fields,

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we're going to store the actual values that we want to represent in our sequence, x0 through xn minus 1, in order.

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And then we're going to use the next pointers to link these all together in that order.

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So the next pointers are what actually give us the order.

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And in addition, we're going to keep track of what's called the head of the list.

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So the data structure is going to be represented by head.

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If you wanted to, you could also store length.

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So this could be the data structure itself.

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And it's pointing to all of these types of data structures.

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Notice we've just seen an array -based data structure, which is just a static array.

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And we've seen a pointer -based data structure.

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And we're relying on the fact that pointers can be stored in a single word,

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which means we can dereference them and we can see what's on the other side of the pointer

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in constant time in our word RAM model.

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In reality, each of these nodes is stored somewhere in the array of the computer.

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So maybe each one is two words long, so maybe the first node is here, maybe the second node is here, the third node is here.

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They're in some arbitrary order.

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We're using this fact that we can allocate an array of size as n in linear time.

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In this case, we're going to have arrays of size 2.

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We can just say, oh, please give me a new array of size 2.

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And that will make us one of these nodes.

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And then we're storing pointers.

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Pointers are just indices into the giant memory array.

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They're just, what is the address of this little array?

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If you've ever wondered how pointers are implemented,

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they're just numbers that say where in memory is this thing over here and in memory they're in arbitrary order.

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This is really nice because it's easy to manipulate the order of a linked list without actually physically moving nodes around,

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whereas arrays are problematic.

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Maybe it's worth mentioning.

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Let's start analyzing things.

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So we care about these dynamic sequence operations.

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And we could try to apply it to the static array data a structure,

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or we could try to implement these operations in a static array.

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It's possible, just not going to be very good.

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And we can try to implement it with linked lists.

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And it's also not going to be that great.

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Let's go over here.

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Our goal is the next data structure, which is dynamic arrays.

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But linked lists and static arrays each have their advantages.

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So let's first analyze dynamic sequence operations.

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first on a static array, and then on a linked list.

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So on a static array, I think you all see, if I try to insert at the beginning of the static array,

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that's kind of the worst case.

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If I insert first, then everybody has to shift over.

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If I'm going to maintain this invariant, that the ith item in the array represents, I guess I didn't write it anywhere here,

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maybe here, static array, we're going to maintain this invariant that a of i represents xi.

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If I want to maintain that at all times, when I insert a new thing in the front,

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because the indices of all the previous items change, I have to spend time to copy those over.

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You can do it in linear time, but no better.

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So static array, insert and delete anywhere,

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cost theta n time.

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Actually, for two reasons.

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Reason number one is that if we're near the front, then we have to do shifting.

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What about insert or delete the last element of an array? Is that any easier?

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Because then if I insert the very last element, none of the indices change.

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I'm just adding a new element.

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So I don't have to do shifting.

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So can I do insert and delete last in constant time in a static array?

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Yeah.

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No, because the size is constant.

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No, because the size is constant.

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So our model is that, remember, our allocation model is that we can allocate a static array array of size n,

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but it's just a size n.

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I can't just say, please make it bigger by one.

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I need space to store this extra element.

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And if you think about where things are in memory, when you call to this memory allocator, which is part of your operating system,

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you say, please give me some, a chunk of memory, it's going to place them in various places in memory.

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And some of them might be next to each other.

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So if I try to grow this array by one, there might already be something there.

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And that's not possible without first shifting.

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So even though in the array I don't have to do any shifting, in memory I might have to do shifting.

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And that's outside the model.

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So we're going to stick to this model of just you can allocate memory.

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You can also deallocate memory just to keep space usage small.

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But the only way to get more space is to ask for a new array.

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And that new array won't be contiguous to your old one. So question?

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So how can you do a dynamic array with a static array?

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What is a dynamic array will be the next topic.

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So maybe we'll come back to that.

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Yeah?

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OK.

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So in a static array, you're just not allowed to make it bigger.

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And so you have to allocate a new array, which we say takes linear time.

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Even if allocating the new array didn't take linear time, you have to copy all the elements over from the old array to the new one

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then you can throw away the old one.

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So just the copying from an array of size n to an array of size n plus 1, that will take linear time.

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So static arrays are really bad for dynamic operations, no surprise.

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But you could do them.

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That's static array.

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Now linked lists are going to be almost the opposite.

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Well, almost.

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So if we store the length, we can compute the length through the array very quickly.

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We can insert and delete at the front really efficiently.

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If I want to add a new item as a new first item, then what do I do?

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I allocate a new node, which I'll call x.

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So this is insert first of x.

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I'll allocate a new array of size 2.

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I'm going to change this head pointer.

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Maybe I should do that later.

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So I'm going to set the next pointer here to this one.

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And then I'm going to change this head pointer to point to here.

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And boom, now I've got a linked list.

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We don't know anything about the order and memory of these lists.

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We just care about the order that's represented implicitly by following the next pointers repeatedly.

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So now I've got a new list that has x in front, and then x0, and then x1, and so on.

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So insert and delete first, at least, are really efficient.

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We won't get much more than that.

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The linked list, insert and delete first are constant time. So that's cool.

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However, everything else is going to be slow.

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So if I want to get the 10th item in a linked list, I have to follow these pointers 10 times.

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I go 0, 1, 2, 3, and so on.

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Follow 10 next pointers, and I'll get the 10th item.

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Accessing the ith item is going to take order i time.

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So get and set at need i time,

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which in the worst case is theta n.

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So we have sort of complementary data structures here.

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On the one hand, a static array can do constant time get at set at.

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So it's very fast at the random access aspect because it's an array.

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Linked lists are very bad at random access, but they're better at being dynamic.

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We can insert and delete at the beginning at least in constant time.

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Now, if we want to actually insert and delete at a particular position, that's still hard because we have to walk to that position.

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Even inserting and deleting at the end of the list is hard, although that's fixable.

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And maybe I'll leave that for problem session or problem set.

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But an easy, so here's a small puzzle.

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Suppose you wanted to solve get last efficiently in a linked list.

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How would you solve that in constant time?

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Yeah.

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Doubly linked list is a good idea, but actually not the right answer.

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That's an answer to the next question I might ask.

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Yeah.

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Store a pointer to the last element.

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Store a pointer to the last element.

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That's all we need here.

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And often a doubly linked list has this.

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They usually call this the tail, head and tail.

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And if you always just store a pointer to the last list, this is what we call data structure augmentation,

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where we add some extra information to the data structure.

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We have to keep it up to date all the time.

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So if we do an insert last or something,

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insert last also becomes easy because I can just add a new node here and update the pointer here.

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Delete last is trickier.

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That's where you need a doubly linked list.

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But whenever I add something to the end of this list, I have to update the tail pointer also.

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As long as I maintain this, now suddenly get last is fast in constant time.

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So linked lists are great if you're working on the ends, even dynamically.

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Arrays are great if you're doing random access and nothing dynamic, nothing adding or deleting being at the ends or in the middle.

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So our final goal for today is to get the best of both worlds with dynamic arrays.

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We're going to try to get all the good running times of linked lists

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and all the good running times of static arrays.

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We won't get quite all of them, but most of them.

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And in some sense, another way to describe what these introductory lectures are about is telling you about how

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Python is implemented.

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So what we're going to talk about next, dynamic arrays, I've alluded to many times,

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but these These are what

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Python calls lists.

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So you don't have to implement a dynamic array by hand

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because it's already built into many fancy new languages for free because they're so darn useful.

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This lecture is about how these are actually implemented and why they're efficient.

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And in recitation notes, you'll see how to actually implement them if all you had were static arrays,

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but luckily we have dynamic arrays so we don't have to actually implement them.

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But inside the Python interpreter, this is exactly what's happening.

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Okay?

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So the idea is to...

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relax the constraint or the invariant,

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whatever, that the size of the array we use equals n,

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which is the number of items in the sequence.

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Remember, in the sequence problem, we're supposed to represent n items.

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With a static array, we allocate an array of size exactly n.

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So let's relax that.

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Let's not make it exactly n.

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Let's make it roughly n.

456

How roughly?

457

You can think about for a while.

458

But from an algorithm's perspective, usually when we say roughly, we mean throw away constant factors.

459

And that turns out to be the right answer here.

460

It's not always the right answer.

461

But we're going to enforce that the size of the array I is theta n,

462

probably also greater than or equal to n.

463

0 .5 n would not be very helpful.

464

So it's going to be at least n.

465

And it's going to be, at most, some constant times n.

466

2n, 10n, 1 .1 times n.

467

Any of these constants will work.

468

I'm going to use 2n here, but there are lots of options.

469

And now, things almost work for free.

470

There's going to be one subtlety here.

471

And I'm going to focus on we're still going to maintain that the ith item of the array represents x i.

472

So this data structure, let me draw a picture.

473

So we've got an array of some size.

474

The first few items are used to store the sequence but then there's going to be some blank ones at the end.

475

Maybe we'll keep track of this.

476

So the data structure itself is going to have an array, and it's going to have a length, something like this.

477

So we're also going to keep track of the length.

478

So we know that the first length items are where their data is, and the remainder are just meaningless.

479

So now, if I want to go and do an insert last, What do I do?

480

I just go to A of length and set it to x.

481

And then I increment length.

482

Boom, easy, constant time.

483

Yeah.

484

How do you know you have enough room?

485

Indeed, I don't.

486

This was an incorrect algorithm.

487

But it's usually correct.

488

As long as I have extra space, this This is all I need to do for insert last.

489

But I'm also going to store the size of the array.

490

This is the actual, this whole thing is size and this part is length.

491

So length is always going to be less than or equal to size.

492

And so there's a problem if length equals size, then I don't have any space.

493

OK.

494

Just add to the end unless n equals size.

495

I'm using n and length for the same thing.

496

So length here is the same as n.

497

That's our actual number of things we're trying to represent. And size.

498

This is great.

499

This is the interface size.

500

This is what we're trying to represent.

501

And this is the representation size.

502

This is the size of my array.

503

These are the number of items I'm trying to store in that array.

504

This is the interface versus data structure.

505

Here's the interface.

506

Here's the data structure.

507

OK, cool.

508

What do I do in the case when n equals size?

509

I'm going to have to make my array bigger.

510

This should sound just like static arrays.

511

With static arrays, we made our array bigger every time we insert it, and that was this linear cost of allocation.

512

We're going to do that sometimes.

513

With static arrays, we had to do it every single time because size equaled n.

514

Now we have some flexibility.

515

We're only going to do it sometimes.

516

It's like cookies are sometimes food, apparently, according to modern cookie monster.

517

I don't understand.

518

But if n equals size,

519

we're going to allocate a new array. of size?

520

Any suggestions?

521

Bigger.

522

Bigger.

523

I like it.

524

Greater than size.

525

How much bigger?

526

Twice.

527

Five things.

528

Five things.

529

Size plus five?

530

Come on, Jason.

531

Trolling me.

532

All right, so there are a couple of natural choices here.

533

One is a constant factor larger.

534

You could use 1 .1 or 1 .01 or 2 or 5 or 10.

535

They'll all work.

536

Or you could use Jason's trolling answer of size plus a constant, like 5.

537

Why is this bad?

538

Yeah.

539

If you like score like six extra three items to that constant, you have to .

540

You'll have to do it again.

541

Again, you'll have to resize frequently. When?

542

Five steps later.

543

So in the original static array, we were reallocating every single time.

544

That's like n plus one.

545

If we do n plus five, that really doesn't change things if we ignore constant factors.

546

Now we'll have to spend linear time every five steps instead of linear time every one step.

547

That's still linear time per operation.

548

It's just we're changing the constant factor.

549

OK, whereas two times size, well, now we have to think a little bit harder here,

550

let's just think about the case where we're inserting at the end of an array.

551

So let's say we do n insert las from an empty array. When do we resize?

552

Well, at the beginning, I guess I didn't say what we do for an empty array.

553

Let's say size equals 1.

554

So we can insert one item for free.

555

As soon as we insert the second item, then we have to resize.

556

That seems bad.

557

Immediately, we have to resize.

558

Then we insert the third item.

559

Now let's draw a picture.

560

So we start with one item.

561

We fill it up.

562

Then we grow to size 2, because that's twice 1.

563

Then we fill it up.

564

Immediately, we have to resize again.

565

But now we start to get some benefit.

566

Now we have size 4.

567

And so we can insert two items before we have to resize.

568

And now we're size eight.

569

And we get to insert four items before we refill.

570

So this is going to resize.

571

And again, resizes are expensive, both because we have to pay to allocate the new array.

572

I drew it as just extending it.

573

But in fact, we're creating a whole new array.

574

And then we have to copy all of the items over.

575

So there's the allocation cost and then the copying cost.

576

So it's linear either way.

577

But we're going to resize at n equals 1, 2, 4, 8, 16.

578

You know this sequence.

579

All the powers of 2 because we're doubling.

580

That is exactly powers of 2.

581

So we pay a linear cost.

582

So this resize cost, the allocation and the copying, is going to be, it's linear each time.

583

So it's 1 plus 2 plus 4 plus 8 plus 16.

584

Really, I should write this as sum from i equals 1 to roughly log n.

585

Log base 2 of n is lg of 2 to the i.

586

If you want a terminus here, it's roughly n.

587

It's actually the previous power of 2 of n or something.

588

OK, but that won't matter.

589

That'll just affect things by a constant factor.

590

What is the sum of 2 to the i?

591

This is a geometric series.

592

Anyone know the answer?

593

Yeah?

594

2 to the top limit plus 1 minus 1.

595

Yeah, so this is the identity.

596

The sum of 2 to the i from i equals 1 to k is 2 to the k plus 1 minus 1.

597

So the plus 1 is upstairs, minus 1 is downstairs.

598

An easy way to remember this is if you think in binary, as we all should.

599

We're computer scientists.

600

2 to the i means you set the ith bit to 1.

601

So here is a bit string.

602

This is the ith bit.

603

This is 2 to the i.

604

The 0 is down here.

605

If I sum them all up, what that means is I'm putting ones here.

606

And if you think about what this means, this is up to k from 0.

607

Sorry, I should do 0 to be proper. If I write.

608

So that's the left -hand side.

609

The right -hand side is 2 to the k plus 1, which is a 1 here, and the rest 0's.

610

So if you know your binary arithmetic, if you add 1 to this, you get this.

611

Or if you subtract 1 from this, you get this.

612

This is why this identity holds.

613

Or the higher level thing is to say, oh, this is a geometric series.

614

So I know you should know this.

615

I'm telling you now.

616

Geometric series are dominated by the last term, the biggest term.

617

If you have any series you can identify as geometric, which means it's growing at least exponentially, then in terms of theta notation,

618

you can just look at the last term and put a theta around it, and you're done.

619

So this is theta of the last term, like 2 to the log n, which is theta n.

620

Cool.

621

Linear time.

622

Linear time for all of my operations.

623

I'm doing n operations here, and I spend linear total time to do all the resizing.

624

That's good.

625

That's like constant each, kind of.

626

The kind of is an important notion which we call amortization.

627

So I want to say an operation takes takes t of n amortized time if,

628

let's say, any k of those operations take,

629

at most, k times t of n time.

630

This is a little bit sloppy, but it would be good enough.

631

So the idea is here, if this works for n or k, to do n operations from an empty array here takes linear time,

632

which means I would call this constant amortized.

633

Amortized means a particular kind of averaging, averaging over the sequence of operations.

634

So while individual operations will be expensive, one near the end, when I have to resize the array,

635

is going to take linear time just for that one operation.

636

But most of the operations are cheap.

637

Most of them are constant.

638

So I can think of charging that high cost to all of the other operations that made it happen.

639

And so this is averaging over the operation sequence.

640

Once every insert last over there only takes constant time on average over the sequence of operations that we do.

641

And so it's almost constant.

642

It's not quite as good as constant worst case,

643

but it's almost as good and it's as good as you could hope to do in this dynamic array allocation model.

644

Let me put this into a table and you'll find these in the lecture notes also.

645

So we have on the top the main operations of sequence interface, which we will revisit in lecture seven.

646

We'll see some other data structures for this.

647

GetAt and SetAt in the first column, insert and delete first, insert and delete last, insert and delete at an arbitrary position.

648

So we've seen three data structures now.

649

Arrays, we're really good at GetAt, SetAt.

650

They took constant time.

651

That's the blue one.

652

We're remitting the thetas here.

653

All the other operations took linear time no matter where they were linked lists were really good at insert and delete first.

654

They took constant time, but everything else took linear time in the worst case.

655

These new dynamic arrays achieve get at and set at in constant time

656

because they maintain this invariant here that a of i equals xi.

657

So we can still do gets and set at quickly.

658

And we also just showed that insert last is is constant amortized.

659

Delete last, you don't have to resize the array.

660

You could just decrease length, and boom, you've deleted the last item.

661

It's not so satisfying, because if you insert n items and then delete n items, you'll still have an array of size theta n,

662

even though your current value of n is 0.

663

You can get around that with a little bit more trickery, which are described in the lecture notes.

664

But it's beyond the we're only going to do very simple amortized analysis in this class

665

To prove that that algorithm is also constant amortized, which it is, you'll see in 046 or you can find it in the CLRS book.

666

That's it for today.

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इस पाठ की शब्दावली और बोलने के सुझाव

यह स्पीकिंग पाठ वीडियो “2. Data Structures and Dynamic Arrays” पर आधारित है। सबसे ज़्यादा दोहराए गए शब्द ये हैं: array, size, insert, constant, item। इस वीडियो में शैडोइंग के लिए 666 वाक्य और 6917 शब्द हैं। बोला गया हिस्सा 49:57 लंबा है। वक्ता लगभग 138 शब्द प्रति मिनट की स्थिर गति से बोलता है, जो शैडोइंग के लिए आरामदायक है। 82% शब्द अंग्रेज़ी के 3,000 सबसे आम शब्दों में से हैं; बाकी शब्द अभ्यास से पहले देख लेना अच्छा रहेगा।

इस वीडियो की मुख्य शब्दावली

वीडियो के 15 सीखने लायक शब्द, उच्चारण और अर्थ के साथ:

शब्द	उच्चारण	अर्थ
static विशेषण	/ˈstæt.ɪk/	गतिहीन, अचल
dynamic विशेषण	/daɪˈnæm.ɪk/	क्रियाशील, गतिशील
interface संज्ञा	/ˈɪn(t)ɚˌfeɪs/	राब्ता, वास्ता
algorithm संज्ञा	/ˈælɡəɹɪðm̩/	अल्गोरिद्म
solve क्रिया	/sɒlv/	सुलझाना, हल करना
lecture संज्ञा	/ˈlɛk.t͡ʃə/	लेक्चर
element संज्ञा	/ˈɛləmənt/	तत्त्व
minus	/ˈmaɪ.nəs/	कम, घटाकर
allocation संज्ञा	/ˌæl.əˈkeɪ.ʃən/	आवंटन
notion संज्ञा	/ˈnoʊʃən/	संकल्पना
arbitrary विशेषण	/ˈɑː.bɪ.tɹə.ɹi/	यादृच्छिक
tail संज्ञा	/ˈteɪ̯l/	पूंछ, पूँछ
analyze क्रिया	/ˈæn.əˌlaɪz/	विश्लेषण करना
ignore क्रिया	/ɪɡˈnɔː/	उपेक्षा, उपेक्षा करना
label संज्ञा	/ˈleɪ.bəl/	लेबल

दोहराने लायक वाक्य

वीडियो के छोटे, पूरे वाक्य जिन्हें आप रोज़ की बातचीत में इस्तेमाल कर सकते हैं:

Are you with me?
Today, we're not doing algorithms.
Why do I say that?
How would you solve that in constant time?
How do you know you have enough room?

इस वीडियो का व्याकरण

वक्ता जिन संरचनाओं का सबसे अधिक प्रयोग करता है, वीडियो के असली शब्दों के साथ:

संरचना	वीडियो में
शर्त वाले वाक्य if + उपवाक्य, will/would + क्रिया — शर्त और उसका नतीजा	If you were a mathematician, you would say
Passive voice be + past participle — ज़ोर इस पर कि क्या होता है, न कि कौन करता है	are supported · can be solved · is called
Relative clauses who / which + उपवाक्य — व्यक्ति या वस्तु के बारे में अतिरिक्त जानकारी	arrays, which is · ram, which is · consecutive, which means
Present perfect have/has + past participle — बीता हुआ काम जिसका असर अभी भी है	We've taken · we've just seen · we've seen

ध्यान देने योग्य उच्चारण

वक्ता 165 बार संक्षिप्त रूपों का प्रयोग करता है, जैसे we're, I'm, don't। इन्हें वैसे ही छोटा बोलें जैसा आप सुनते हैं।

“th” की ध्वनियाँ: algorithm /ˈælɡəɹɪðm̩/, bother /ˈbɑðɚ/, theta /ˈθiːtə/, arithmetic /æɹɪθˈmɛtɪk/, mathematician /ˌmæθ.(ə.)məˈtɪʃ.ən/
“sh” और “zh” की ध्वनियाँ: allocation /ˌæl.əˈkeɪ.ʃən/, notion /ˈnoʊʃən/, efficient /ɪˈfɪʃənt/, representation /ˌɹɛp.ɹə.zɛnˈteɪ.ʃən/, assumption /əˈsʌm(p).ʃ(ə)n/
लंबे शब्द — बलाघात सही रखें: allocation /ˌæl.əˈkeɪ.ʃən/, arbitrary /ˈɑː.bɪ.tɹə.ɹi/, geometric /ˌd͡ʒiː.əʊˈmɛt.ɹɪk/, representation /ˌɹɛp.ɹə.zɛnˈteɪ.ʃən/, consecutive /kənˈsɛkjʊtɪv/

हिंदी भाषियों के लिए मुश्किल ध्वनियाँ:

/v/ और /w/ का फ़र्क़ — /v/ में ऊपर के दाँत निचले होंठ को छूते हैं, /w/ में होंठ गोल होते हैं: sequence /ˈsiː.kwəns/, solve /sɒlv/, versus /ˈvɝsəs/, whereas /(h)wɛɹˈæz/, consecutive /kənˈsɛkjʊtɪv/
/z/ — इसे “ज” (/dʒ/) न बनाएँ: whereas /(h)wɛɹˈæz/, analyze /ˈæn.əˌlaɪz/, representation /ˌɹɛp.ɹə.zɛnˈteɪ.ʃən/, afterwards /ˈæf.tɚ.wɚdz/, physically /ˈfɪz.ɪ.kli/
शब्द की शुरुआत में s + व्यंजन — पहले “इ” न जोड़ें: static /ˈstæt.ɪk/, specify /ˈspɛs.əˌfaɪ/, string /stɹɪŋ/, straightforward /ˌstɹeɪtˈfɔː(ɹ)wə(ɹ)d/, spell /spɛl/

इस वीडियो से अभ्यास कैसे करें

पहले बिना बोले पूरा वीडियो एक बार सुनें और जो शब्द नहीं आते उन्हें लिख लें।
सामान्य गति पर एक-एक वाक्य की शैडोइंग करें और हर वाक्य तब तक दोहराएँ जब तक आपकी लय वक्ता से न मिल जाए।
अपनी आवाज़ रिकॉर्ड करें और मूल से तुलना करें, ख़ासकर static, dynamic, interface जैसे शब्दों पर ध्यान दें।

शैडोइंग तकनीक क्या है?

शैडोइंग (Shadowing) एक विज्ञान-समर्थित भाषा सीखने की तकनीक है जो मूल रूप से पेशेवर दुभाषिया प्रशिक्षण के लिए विकसित की गई थी। विधि सरल लेकिन शक्तिशाली है: आप मूल अंग्रेज़ी ऑडियो सुनते हैं और तुरंत इसे ज़ोर से दोहराते हैं — जैसे वक्ता की छाया 1-2 सेकंड की देरी से। शोध से पता चलता है कि यह उच्चारण सटीकता, स्वर, लय, जुड़ी हुई ध्वनियाँ, सुनने की समझ और बोलने की प्रवाहशीलता में काफ़ी सुधार करता है।

शैडोइंग तकनीक: पूरी चरण-दर-चरण गाइड पढ़ें →

शैडोइंग अभ्यास: 2. Data Structures and Dynamic Arrays - वीडियो के साथ अंग्रेजी बोलना सीखें

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इस वीडियो से अभ्यास कैसे करें

शैडोइंग तकनीक क्या है?