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>> MARK GROZEN-SMITH: Hi, I'm Mark
Grozen-Smith, and this is Quicksort.

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Just like insertion sort and bubble
sort, Quicksort is an algorithm for

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sorting a list or an array of things.

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For simplicity, let's assume that those
things are just integers, but

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know that Quicksort works for
more than just numbers.

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Quickstart is a bit more complicated
than insertion or bubble, but it's

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also much more efficient
in most cases.

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Wait a second.

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Did he just say "most
cases," not "all"?

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Interestingly, no.

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Not all cases are the same.

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Don't worry about this detail if you
haven't seen big O notation yet, but

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Quicksort is an O (n squared) algorithm
at worst, just like

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insertion or bubble sort.

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However, it typically acts much more
like an old analog m algorithm.

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Why?

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We'll get back to that later.

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But for now, let's just learn
how Quicksort works.

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>> So let's walk through Quicksorting this
array of integers from smallest

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to largest.

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Here we have the integers 6,
5, 1, 3, 8, 4, 7, 9, and 2.

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First, we pick the final element of
this array-- in this case, two--

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and call that the "pivot." Next, we
start to look at two things--

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one, the lowest index, which I'll refer
to as staying to the right of

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the wall, and, two, the leftmost
element, which I'll call the "current

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element." What we're going to do is
look all the other elements, other

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than the pivot, and put all the elements
smaller than the pivot to the

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left of the wall and all those
larger than the pivot to the

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right of the wall.

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Then, finally, we'll drop the pivot
right on that wall to put it between

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all the numbers smaller than it
and all the numbers larger.

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>> So let's do that.

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Pick up the 2, put the wall at the
beginning, and call the 6 the "current

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element." So we want to look at
our current element, the 6.

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And since it's larger than the
2, we leave it there to the

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right of the wall.

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Then, we move on to look at the 5 as our
current element and see that this

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is, again, larger than the pivot, so we
leave it where it is on the right

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side of the wall.

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We move on.

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Our current element is
now 1, and-- oh.

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This is different now.

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The current element is now smaller than
the pivot, so we want to put it

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to the left of the wall.

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To do this, let's just switch the
current element with the lowest index

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sitting just to the right of the wall.

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Now, we move the wall up one index
so the 1 is on the left

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side of the wall now.

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>> Wait.

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I just mixed up the elements on the
right side of the wall, didn't I?

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Don't worry.

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That's fine.

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The only thing we care about for now
is that all these elements to the

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right of the wall are bigger
than the pivot.

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No actual order is assumed yet.

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>> Now, back to sorting.

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So we continue on looking at
the rest of the elements.

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And in this case, we see that there are
no other elements less than the

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pivot, so we leave them all on
the right side of the wall.

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Finally, we get to the current element
and see that it is the pivot.

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Now, that means that we have two
sections of the array, the first being

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small on the pivot and on the left side
of the wall, and the second being

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larger than the pivot to the
right side of the wall.

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We want to put the pivot element between
the two, and then we'll know

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that the pivot is in its right
final sorted place.

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So we switch the first element on the
right side of the wall with the pivot,

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and we know the pivot's
in its right position.

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>> We then repeat this process for the
subarrays left and right of the pivot.

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Since the last subarray is only one
element long, we know that's already

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sorted because how can you be out of
order if you're only one element?

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For the right side of the subarray, we
see that the pivot is 5, and the wall

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is just left of the 6.

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And the current element also
starts out as the 6.

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So 6 is greater than 5.

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We leave it where it is on the
right side of the wall.

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Now, moving on, 3 is less than 5.

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So we switch it with the first element
just right of the wall.

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Now, I moved the wall up one.

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Now, moving on to the 8.

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8 is greater than 5,
and so we leave it.

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The 4 is less than 5, so we switch it.

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And on.

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And on.

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>> Each time we repeat the process on the
left and right sides of the array. we

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choose a pivot and do the comparisons
and create another level of left and

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right subarrays.

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This recursive call will continue until
we reach the end when we've

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divided up the overall array into
just subarrays of length 1.

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From there, we know the array is sorted
because every element has, at

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some point, been a pivot.

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In other words, for every element, all
the numbers to the left are lesser

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values and all the numbers to the
right have greater values.

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>> This method works very well if the
value of the pivot chosen is

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approximately in the middle
range of the list values.

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This would mean that, after we move
elements around, there about as many

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elements to the left of the pivot
as there are to the right.

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And the divide-and-conquer nature of the
Quicksort algorithm is then taken

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full advantage of.

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This creates a runtime of O (n log n,)
the n because we have to do n minus 1

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comparisons on each generation and log
n because we have to divide the list

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log n times.

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However, in the worst cases, this
algorithm can actually be O (n

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squared.) Suppose on each generation,
the pivot just so happens to be the

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smallest or the largest of the
numbers we're sorting.

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This would mean breaking down the list
n times and making n minus 1

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comparisons every single time.

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Thus, o of n squared.

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>> How can we improve the method?

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One good way to improve the method is
to cut down on the probability that

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the runtime is ever actually
o of n squared.

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Remember this worst, worst case scenario
can only happen when the

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pivot chosen is always the highest
or lowest value in the array.

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To ensure this is less likely to happen,
we can find the pivot by

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choosing multiple elements and
taking the median value.

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>> My name is Mark Grozen-Smith,
and this is CS50.

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>> For simplicity, let's assume that those
things are just integers, but

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know that Quicksert--

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Quicksert?

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Sorry.

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>> Here we have the integers
6, 5, 1, 3, 8, 4, 9.

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>> SPEAKER 1: Really?

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>> SPEAKER 2: Don't stop there.

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>> SPEAKER 1: Really?

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