so now we're going to look at 4 examples of solving a particular problem
and this will demonstrate the speed of each
and the variation of how we solve it so I'll give you an idea what we're gonna be
doing the rest the class as we look at different solutions for a problem
and we are also going to show you one of the pitfalls of
solving a problem that you have to look out for. So are talking about anagrams
here's an example for him in a gram hearten earth I
and here's another one tight on time typhoon and so what is an anagram
you have two strings and its
you could take the first ring and rearrange it in some way and get the
second-string
so struck in string the first ring have alt same characters
and they're just a rearrangement up the are
the months from one screen to the other are the strings don't have to be actual
english words but
usually when you're solving puzzles you'll give you'll be given two words
for technically that's an n-gram
so we look at different approaches and look at
what are how much it cost us to do each approach
so the first lucien he calls checking of and the idea is that we have
two strings so we're going to are
look at the letter H in the second string and check of what we find it
so when you find it we're gonna check it off and we can check it off
rely replacing the age with a non now we can't do that in a string
the two mutable mean we can't modify anything in this crying
so what we're going to do is convert these from strings to list
and mister not immutable so we can actually replaced H
in the first with non an so we go to the
animate check often find the EU are the firstly we found
and then we find today evening on check out of and so on
and if we're all done we've checked everything of your
and we've also finish going through this whole spring here we know it's an
anagram
now we're gonna show you two versions though version the book author did
and the version I've done have tried to simplify some things a little
user code understand and we're gonna take are some advantages of things
for small if these two strings are not equal
they automatically or not anagrams that's what something we can test
immediately industry turn
we find that they're not equal are so let's go ahead and see
the author solution and
are we go back and do so what you gonna do is you gonna
pick the H and then for this age you're looking through the string
and then you gonna pick the and for the economy looking
string so you can see that we're gonna have a loop
but I loop through the first string one at a time and then
inner loop so that when the first ring is on the H it'll start
looping through all the letters are the second-string so that's basically what
he does here
so he had she all these are gonna define an interim solution or two strings
so here this is where it converts a string
to a list and this is going to be useful to remember how to do
anything that's a string you can convert to a list to the individual characters
sold he's gonna you some variables cue track up where he is and each list so
position one more point to
where we are and the s1 worst
an position to a point where we are and the
s2 or the a-list and we only have two
at nine today with us we don't have to are convert s 12 at worst
and we have a flag that so it's still okay
means we're still looking for an anagram and so while
position one in the first ring is lost in late to the string
and still okay to gonna reap set the pointer position
are two in the second string 20 search found two faults we have another flag
we're gonna set when we actually find it and then while
position to pointing to the second string is less than a link to that list
and not found we're gonna checkers
the is the screening at position 1 equal to the list
position to after two years we want to
such found a true
and when we sat down the true its gonna pack go back up to the while an hour
cause the loop to terminate early
if it's not equal then we increments
opposition to a look at the next character in the list
to see you if the what we're looking for from the first list is in the second
position and so on
so this will loop through looking at the second-worst are
and here's the outer loop which is looping through the first
screening now once we've done through all the way through the list we check
found if we never found what you're looking for
are families know if we did find what you're looking for
we set position where we found it 29 so that's gonna market
or check it off if we never found it we set still OK to faults
that means we went through are this whole
less time we never found a particular character that means it can't be an
anagram
so still a case at faults its gonna cause this outer loop to quit
I and then finally we increment
the outer blue we're from an the pointer where we are in the first ring by one
and repeat outer if the outer loop
still okay becomes faults or weaved run out screaming
check we exit outer loop and wait to return still okay
nets can return to your fault tan and brown are
your cell the way I wrote it so on going to in a gram 1
so the way I wrote it for such check if the link to the two strains are not
equal
and if it is I returned faults an
10 I set in the quarter to the link to the first ring
and I make a new list I called s2
worst so it s too was a list and it for each character in s1
then for each hi in the range
%uh the link to the STU so I will be an index into the second
list excuse me
so if the character from the first
are string is equal to bite
position at the second-worst then we market of
and then we r
document and buy one so Ennis keeping track
old are started out with the link to
the second-string and so every time we str actor were marking off if we marked
off everything
in this second-string we know we're done
and we found an anagram so that's going to be our flag which Jack
I when you're all done so we don't mean it don't actually have any flags
are but we actually found the character we can break out of the loop
sold break you members a special key word in Python
works just like in Java and C plus plus which will break out the inner loop
and go to the next character the outer so what we gone through all the
characters are the outer loop
are then we just return it in a zero it's an anagram
so checks return an equal 0 if it's not zero we didn't check turning off
and so it's not a nanogram now we're gonna see more
this this is testing so what I did as I made a top of tuples
is called the test cases so each tuple
that's inside as a whole list to top holes are three things each tuple
Thursday word to check and a first ring the second string and what we should
expect it to be
so this is actually to anagrams these are anagrams
ABC with an empty string is not an anagram
to empty strings were gonna know just that they're true
now this particular case it turns out the authors
are I think daughters example fails at it or
fails at one of these were you have an empty string are so
you don't have to have this work all the time for empty strings
is just doesn't make sense but it does work on our scores we
test to the springs or he call and with check-in is early and so will be
so the sole return true I and we have a bunch other test cases
and so what we do is we loop through this is a special
our wave looping through topple so we have this is a list of tuples
so each particular piece %uh this is a tuple and so we're using that assignment
where you can list
three variables on the left so inside the top all
s1 as to EU will be assigned to these three things in order
so this is in the wave using a for loop on a plane out
and they were using the assert statement and what the research statement doesn't
play it on
is it's always a cert and then something is true or false
and then you can have optionally a string so I have a format string here
which is %uh like an error message
solar search you can add a bunch a search and if your program is
it all these assertions of this are true
assert doesn't do anything you won't print anything it doesn't stop your
program
so this is a lot of times how your test a program and we'll see this one you do
fraction
arm that when you did fraction the previous chapter is asserts
so if our anagram solution is equal to what
is expected the thats whats asserting
if it's not it it will output in error message
quit the program and their messages space layout putting what the two
strings returned were
so you know what data failed the
test so stock up a little bit about big old this
because we're were stepping through every character the first ring
and then we have another look to the stepping sequentially through every
character
the second-string analyst art this is two nested for loops
this one's going from one day in and this is going to one and where the end
is the size of the problem
so I put their own notes here this is if you counted how many assignments is one
assignment here
there's one assignment to make a new list
you could argue that's actually an operation are
then there's an square in squared assignment are
this 102 and it depends on how far idk checks
how far gets analyst can we have in
squared here and these are all in square because you're inside at the
nested who and then there's one assignment when rolled on
and I didn't even count that so this this I've put a limitation
as to what polynomial all this adds up to
but the main thing you'll see is the biggest term I use upchurch for power
so we have if we simplify this it's to an
squared plus and plus two so the
property predominant turner's in squared so this is all 11 squared
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