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Oh, right.

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So I think this process was not an easy one for you, but still, if you're here, I want to thank you

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and I want to say that you are doing great because even coming to the solutions video is not an easy

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task after these.

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Not so easy exercise.

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So let us get started.

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All right.

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And what I want us to do, first of all, is basically to try to think even how would you approach this

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exercise?

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So let's say that we have here, let me get my pen.

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OK, so let us take a look at this example.

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Four, three, eight.

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OK, so since we know that we are going to use a recursion function that probably probably there is

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going to be some recursive calls.

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Right.

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For or recursive instances.

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So let's say there is an instance for an equals two, four, three, eight and then an equals two,

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four, three and then an equals to four.

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Basically, that's it.

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OK, I'm just I'm just trying to like to go over the steps and over your.

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Possibly thoughts that you had during the solution, so are and equals two, four, three, eight and

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equals two for a three and and equals to four.

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So we can say that whenever we reach some point when and he's less than 10, OK, which means it's going

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to be it's going to be one digit number, then in this case, we have to like to think what should be

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returned.

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So that's very interesting.

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Guys, what do you think should be returned here?

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Should it be if if that's the case?

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Because because I think that a lot of you guys thought of this way of and what will happen if and is

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less than 10, which is something that you would have used for a base case, what do you think should

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be returned in such a case?

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Should be returned just the value of one, OK, or the value of zero, basically saying we said previously

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that we will return one if every digit and that an even possession has an even value as well as every

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digit in the opposition will have an odd value.

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But here we have like just one one value of one digit.

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So how can we say is this position is.

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Is this position is more positive, this position is even or odd, so what do you think?

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And to be honest, I think it's kind of hard to assess if these four here was at an even or odd position,

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since there may be also another number, like four or three, four, four, three, eight, OK, and

40
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it will basically it's a lot like this, but yeah, four four, four, three, eight.

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And it will basically look something like this.

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Right.

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So it will look silly.

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Not not like this.

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So for three, eight, four, three, eight and let's say I don't know, let's say five for example.

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So this will be just also the recursion calls for and divided by ten for every recursive call.

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So what I want to say by these is that also for at least number and also for this number, the value

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of four can be at an even index like here, or it can be in an odd index like here.

49
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Right, because zero.

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One, two, three.

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So.

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The assumption of if is less than 10, then return one or zero, just based on the fact that you have

53
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one digit is not the correct way to go.

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00:04:21,900 --> 00:04:23,650
So what do you think?

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How should we basically approach that?

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And one of the things that I suggest to you to understand is the fact that, first of all, for every

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natural number you may have for the end itself, OK, for the end itself, that's the very that's the

58
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catch here, let's say, or the basic trick.

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So for every M, there are only two options.

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OK, two options either.

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Either the number of digits is even or odd.

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OK, so even.

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We're on.

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Number of digits, OK, so every number is compromised of digits, so every number goes like this,

65
00:05:09,950 --> 00:05:19,100
like digit one digit too, and so on up until some digit M right every number, this may be one digit

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number, two digits number and so on.

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And the number of these digits, maybe either odd or even.

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OK, so that's the first step that I want you to understand, because that's going to be crucial and

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very important for our solution.

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So even and odd.

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So even and odd does not mean that there is something about the positions that that's just the dust.

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That just means that there are even even digits or digits.

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So that example here, like twenty six or four, eight, seven, five, I don't know, two digits,

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four digits, even number of digits.

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Odd will be like three five seven five.

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I don't know, something like that.

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OK, three digits, one digit odd number of digits.

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So let's just on droid here that everything will be also clear to you.

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Number of digits.

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Awesome.

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So that's the number of digits, number of digits, and now what I want us to do is basically to use

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the same approach that we used previously by just taking the end and to call these recursive calls.

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OK, but these recursive calls are going to have some catch, OK?

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On every iteration, I want us to divide the value by a hundred, OK, and to test two.

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Basically two are things the value on the rightmost value and the value the other value that I will

86
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just show you in a bit.

87
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OK, the rightmost value and one to each left and we will have two scenarios here.

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First scenario is if and is less than 10.

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OK, meaning we got one digit and if and is less than a hundred.

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OK, so these are going to be our two base cases.

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And why is that why do we have to base cases?

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Because whenever we will have and deal with our values that have like that have even numbers of digits,

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then in this case, every time that you are going to divide it by 100.

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OK, then let's say I don't know, let's say four four eight seven five.

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You divided by a hundred, you get like forty eight.

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So you will always be like in this region.

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OK, for all the values that have like a number of digits, even every time you divide by 100, you

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will always fall here.

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Either way, it's going to be like, no, like these two six, five, four, three, two.

100
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You divided by a hundred and then you divide it by also a hundred.

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And then this value is going to be here.

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OK, you are never going to reach these base case where you simply have this if and these less than

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10 never for even for one number of digits equal to even on the other hand, you are always going to

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reach these place for numbers that have an odd number of digits.

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So in this case, three will reach here five seven five.

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When you divide it by a hundred, you're also going to remain with five and you are going to reach this

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place.

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So this fact, the differences between these two base cases is actually used to distinguish between

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two cases when we have a number of digits and even number of digits and an odd number of digits.

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And why is it so useful?

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That's useful because it will allow us during the process to know in advance, OK, to compare basically

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these two values, the rightmost and one on its left, and simply to build our way up once we know the

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position.

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OK, so here we will know that if we if any less than one hundred, let's say it will be like four eight.

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OK, then in this case we know to check that both of them satisfy the condition, if so, return one,

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otherwise return zero.

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And then we will have like on the way up four eight seven five and we will check this out and these

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out and we will know that this will be always at an even position and this will always be at an odd

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position.

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OK, so once again, that's not so trivial to grasp at first.

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OK, I'm going to write the code and then we are going to go over this visualisation process again,

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OK, with a few examples to make sure that everything is clear to you.

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So don't worry, you hopefully will understand everything that goes behind the scenes.

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OK, so now let us start writing of the solution, guys.

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OK, so let's just make sure that I'm recording this.

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Yes.

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And then I'll start with the solution.

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Let's remove everything from here from the screen.

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Awesome.

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And let's go here and let's call our function, first of all, the type of the function.

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What should it be?

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Let's stay with an integer.

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Right.

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Because the two optional values that can be returned from this function are simply either zero or one.

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All right.

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So.

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And let's call it I don't know, I don't know even the or I don't know, um, even on Phonic, let's

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call it for now and get some value or some natural number of him.

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Awesome.

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So we said that we are going to treat mainly to base cases, OK, to base cases.

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The first one is when M is less than 10.

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And if that's the case, let us see if OK if no if these and modular two equals to zero, if that's

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the case, if it a one digit one digit number, OK, one digit number, that simply means that we were

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working with on some end value that have in the first place.

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And the first recursive call, an odd digit, an odd number of digits, then it means that this digit

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OK, this digit right here is going to be at what position at and what do you think.

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And even position, right, because again, here is a simple example, four, three, eight, and if

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we divided by 100, we simply get these four.

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So if we will take a look, these four will be at an even position always.

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Right?

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Right.

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Because every time we are going to divide by 100, it's going to be at an even position.

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Awesome.

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So if and can be divided by two without the remainder, that means that the value is even also.

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And if that's the case, we can assume that we will return one because the condition is satisfied,

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otherwise we will return zero because the condition is not satisfied, meaning we have one digit and

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its position is even, but the value cannot be divided by two without the remainder.

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That means that the final result should be zero because the condition is not satisfied.

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This condition here right for returning one.

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Awesome.

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So now let us proceed for the second base case.

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And these base case is basically something like this if and is less than one hundred, meaning it is

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a two digits number.

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OK, and if that's two digits.

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No, that's awesome.

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Right.

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Because that means we simply are we simply we simply we simply.

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What does it mean?

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What do you think, if anything, less than one hundred.

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What does it mean then.

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It means we are we remain with just two digits, OK.

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And it will look like we don't have an example here.

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So let's say we simply will also make another example.

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We will make another example, its example three, and it will be like this.

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So five here.

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So did it.

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And also we will use also like something like this.

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OK, so position zero, position zero will have five.

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Position one will have eight.

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Position two will have three and position three will have four.

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OK.

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So what do we have here?

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What do we have here if AMM is less than 100 hundred, meaning we divided this time by a hundred, we

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got forty three, then in this case, we know that we will have two digits.

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OK, do digits four and three.

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OK, that's the base case we are planning to reach.

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So four and three and this three lies on position even and these four will lie on position on OK so

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do digits.

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Are the rightmost digit at even position.

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Right.

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Even pause and leftmost, leftmost or left digit is at odd position.

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OK, you see, you see what we're doing here so we know we will have two digits and the rightmost digit

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that can be accessed by using these and modulo 10 can be accessed.

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And it will it is going to be at an even position and the one on its left is going to be at an odd position.

195
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All right, so OK, if that's the case, we are going to see the following if an Modula 10, which is

196
00:16:19,540 --> 00:16:24,720
their rightmost digit, can be divided by two without a remainder.

197
00:16:24,880 --> 00:16:28,420
OK, if it can be divided by two without the remainder.

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00:16:29,020 --> 00:16:38,710
And also if we take the leftmost digit in such a case and divided by two without a remainder or basically

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with the remainder of one.

200
00:16:41,560 --> 00:16:51,870
So if that happens, that will mean that the rightmost digit at the even position is the value is even

201
00:16:52,150 --> 00:17:03,190
and also the left value in between these two digits is an odd value, has an odd value.

202
00:17:03,190 --> 00:17:06,430
And it's also, as we said, at an odd position.

203
00:17:06,790 --> 00:17:10,240
If that's the case, what do you think we should return?

204
00:17:10,630 --> 00:17:13,030
We should return one.

205
00:17:13,570 --> 00:17:18,670
OK, otherwise, of course, we can return zero.

206
00:17:20,320 --> 00:17:21,230
Is that clear?

207
00:17:21,700 --> 00:17:30,130
OK, we just tweeted the two base cases, so that was basically case one base case still in the differences

208
00:17:30,130 --> 00:17:34,510
between them is that these base cases going to be reached.

209
00:17:34,680 --> 00:17:38,170
OK, if we correctly build the following line of code.

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00:17:40,090 --> 00:17:51,070
If we are, it's going to be reached only for numbers that have been in an odd number of digits.

211
00:17:51,070 --> 00:17:58,270
And these base case is going to be reached for numbers that have only on an even number of digits.

212
00:17:59,830 --> 00:18:00,240
Whoo!

213
00:18:00,490 --> 00:18:02,480
All right, so let us proceed.

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00:18:02,530 --> 00:18:09,380
And now what we are going to do is basically to ask the following question.

215
00:18:09,700 --> 00:18:17,500
So that's going to be regarding our, um, our recursive call.

216
00:18:17,890 --> 00:18:20,850
So our recursive call is going to be very simple.

217
00:18:21,220 --> 00:18:29,350
We are going to ask we are going to ask if an modula to or basically Modula 10.

218
00:18:29,350 --> 00:18:42,370
Modula Two, which means the rightmost digit, if it is and even digit and also if this digit of digit

219
00:18:42,370 --> 00:18:52,840
on on its left, OK, one on its left, so much about 10 if this digit the rightmost digit and also

220
00:18:53,020 --> 00:19:00,070
the one digit on its left, OK, and divided by 10 modulo 10 can be divided by two.

221
00:19:00,940 --> 00:19:08,410
OK, if that's the case, OK, we are going to assume that if that's the case, we are going to return

222
00:19:08,410 --> 00:19:09,220
the result.

223
00:19:09,950 --> 00:19:14,320
OK, four are these even all function.

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00:19:14,590 --> 00:19:18,020
Four and a hundred.

225
00:19:19,210 --> 00:19:24,640
OK, so that's very not so intuitive else return zero.

226
00:19:26,360 --> 00:19:33,860
OK, so that's not so intuitive to to grasp at first, but first of all, what I want you to understand

227
00:19:33,860 --> 00:19:43,850
is the distinguished distinguish distinguish we've made between numbers with odd and even number of

228
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digits.

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00:19:44,840 --> 00:19:51,710
And now regarding these recursive call, I think the best way to understand it is simply to use some

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drawing and to use some examples.

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OK, so let us start let me get my pen and let's start drawing.

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OK, so.

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Let's say that we are going to start with this third example, so it's going to be like four, three,

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eight five, OK, this condition is false, this condition is false.

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And we are going to ask if five in this case, that's five.

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If five is even, is there even what do you think?

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Is it even know?

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It's not even.

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But we are checking the even position.

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We are checking the even position.

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OK, that's the rightmost element, if that's the case.

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But it's not, then we simply go to the else and we return zero.

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OK, so the result for these will be zero and it kind of makes sense since you can see here and even

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position but an odd value.

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So that's basically it for this example.

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So that's how you find it.

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Now, let us proceed to another example.

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So let's see what it is.

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Let's go a little bit up.

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So four, three, eight.

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OK, so four, three and four and eight.

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OK, so now we are going to let's just remove that so we'll see everything on the screen.

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Also the code.

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OK, so now we are going to ask the following question.

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Condition, false condition, false.

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So if the rightmost digit is even.

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Yes, it is.

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And one on its left, which is three is odd.

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Yes it is.

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Then we are going to return all four and divided by one hundred, which is four.

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OK, so this result basically depends on what it's going to return us for even often.

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Four, four.

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And if we are going to come here, this is the base case related right to numbers with just odd number

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of digits.

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Then in this case, this condition is true.

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Four is less than 10.

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If four can be divided by two without the remainder, that is true.

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The value is even and it was at an even position, then we are going to return one.

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And also from here we are going to return.

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One final result is one.

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All the conditions, conditions are satisfied exactly as we explained and expected.

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And now let us see the final example.

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OK, I hope everything is clear.

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Ask me if you have still any questions.

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Let us see the final example.

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What was it?

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What was it?

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Three six three six four three five.

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OK, so.

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That's it, three, six, four, three, five.

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OK, so let us check let us check if and Modula 10, which is the rightmost digit, can be divided by

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two without a reminder ups.

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No, it can be so return to zero.

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And there you go.

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Basically, you return zero right from the start.

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So these are this is it about the examples that I wanted to show you.

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Of course, there are a lot of additional examples that we can use, but I think I've covered at least

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a few of them to make it more clear to you.

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And now I want to simply summarize the steps that we used to solve this exercise.

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So.

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Step number one was to distinguish between two main scenarios, scenario number one was like to understand

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that there should be some sort of different treatment and different solution for situations when you

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have an odd or even number of digits in a given number.

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OK, so that was the first thing we had to take care of.

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And we simply got got got it taken care of at this step.

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Of course, if you see it for the first time, this kind of exercise, it's not so easy to grasp and

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it's not so trivial to understand the separation between assuming that if there are going to be odd

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number of digits, then if you divide it by 100 every time and you reach these kind of time that the

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number that digit is going to be at even position.

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And here you will have two digits and, you know, like it's very hard at first time.

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But now, since you kind of know the basics in the approach, it should be easier for you to solve or

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exercises of a similar concept and of a similar complexity.

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OK, so that's about it.

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And the first place then we have like to understand how to make the recursive call.

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So we understood why we're using and divide it by a hundred every time that it will be able to distinguish

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for us two cases like here and every time we don't care if we have like numbers of an odd number of

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digits or an even number of digits, we don't care about it since we start from there.

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Right, OK.

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And when we start from there, right.

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Every time we will see the rightmost element, it should have a value that is even and one only two

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left should have a value that is odd.

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Otherwise there is no sense to make any recursive calls and we will simply assume that the return result

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should be zero.

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OK, so every time we we we since we divide by 100, every time we are going to start like from from

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position zero and then from position to and from position four and so on and so forth.

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So that makes sense.

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OK, and the final distinguish between the two based cases will be handled in these kind of sections.

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Who so guys, this exercisable is not an easy one.

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I hope you've got a hold of it.

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If you still have any questions, feel free to ask them.

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And for those of you that know the answers to the questions being asked, also, we kind and try to

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help your associates and your friends if they have, like, questions that you can address and you can

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answer very accurately, since that's also kind of a good learning process when you try to explain it

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to someone else.

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And yeah, this is it for these Medio guys.

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My name is Vlad.

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This is Alphatech.

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I hope you like the course so far and until next time, I'll see you then.

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Oh, and by the way, don't forget to let me know what you think of the video and maybe to leave some

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review that I will know that the process is going exactly as planned.

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Or at least I hope so.

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Yeah.

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Bye bye.