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Hi, everyone.

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So in the last video, we discussed this question, find the diameter of a binary.

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So in this video, we want to discuss a better approach, better approach in terms of time, complexity.

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So this is how we will utilize our resources.

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This is the left subtree and this is the right Sabry.

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So what will do so rude will ask the left, hey, find out the height and the diameter.

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Solarcentury will determine its height and it will dondi diameter.

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Now Ruth will call on the right subtree.

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Hey, right.

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Find out your height and diameter.

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So I have probably done the height and the diameter so we know what is left, right and left diameter.

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We know what is right height and right diameter.

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So can we find out overall height and overall diameter?

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So what will be the overall height?

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So overall height will be one plus maximum of left and right.

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Right.

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So this will be one plus maximum of left and right.

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Right.

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And can we find out what is the overall diameter?

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So overall diameter will be.

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So I have three options.

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So I take maximum of what is the first option left out plus right.

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So first option is left out.

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Plus right eight.

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Second option is left diameter.

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Third option is Radiometer.

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So we can find out the overnight and we can find out the overall diameter, so root will return overnight

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and overall diameter simple.

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So that is our approach.

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So first, let's discuss the base case.

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So case will be traditional, so if Rudisill is the first step.

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So if traditional, but without zero vertical diameter to zero, now the second step is basically hypothesis

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step or the induction step.

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So in this step, we will be able to find out the left, right, left, diameter, right out and right

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diameter.

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And what is the third step?

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So third step is basically finding out the overall height and diameter and the steps or this step is

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basically the third step, finding the overall height and the overall diameter.

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So what we want to do, so we want to return to things, I want to return to things so we can create

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a class.

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So I want to return a pair of these two things so I can go to class, Gaspare.

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First will be debate in Detroit, and second thing is Daym.

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So I know what the function will be up here, and let's say the name of the function is find out the

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height and the diameter, it will take root as input and it will return dippie.

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So it will be done up here and it has two things height and the diameter, or if you want to make the

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call for the generic so I can create glass like this glassware.

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So instead of this height and diameter, I will call it first and second.

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I am calling it first, I am calling Dimetapp Second.

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So if you want to make the class, if you want to make the class of beer generic so you can do also

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like this in first and second, but there is no need to do all this thing because we have Invicta per

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class.

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So we have inbuilt Berglas in C++ and this class has implemented using template.

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So this is the code for the Beltway class, so instead of integer first, it is the first and similarly

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it is B second, and then you will create the object of this class.

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You will give the value of t, you will give the value of.

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So let's write the code.

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So this is our previous quote, let's find out everything or let's reset our good.

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So I'm resetting my code, so what I want to do.

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They don't have the function will be their superiors and world class and C++, since it is implemented

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using template.

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So the first value height is going to be integer.

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The second value diameter is also integer.

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And let's say the name of the function is height diameter.

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The name of the function is higher diameter.

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So it will take.

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Input.

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And it will be done up here.

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And finally, what we will do so here, we will call this function.

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So let's call this function.

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Hi, Daym.

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We will pass it, Ruth.

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So dysfunction and it does appear so let's create a period.

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Fear of indigenous communities are heightened diameter, let's name it be.

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And what we need to return, so we need to return integer and I want to return diameter.

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So what is diameter?

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Diameter is the second value.

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So I am assuming they do this second.

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So I will write P Dot.

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Second.

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So what I'm doing here, so I'm calling this function Dimitar, I'm giving your route so this function

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will return me what is the height of my tree and what is the diameter of my tree?

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So it is a stunning integer integer.

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So this is an multiclass Glaspie integer first and indeed second.

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So why integer, because you are giving Integer an advantage, because you are giving integer and what

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you need to do.

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So I'm assuming first is right and second is diameter, and this function returns are daym, so I am

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returning Bédard second.

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So I am returning the diameter value and now we need to write the code for this function.

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So let's write the code.

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So basically, if traditional.

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So if traditional, my height will be zero, my diameter will be zero, and I need to return a pair.

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So let's create a pair.

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Fear of contagion combined.

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Let's name it be so what out first with our first sight, Heidi?

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Zero.

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What is that second winner is diameter, diameter, zero, and then you can be so then we are turning

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up here.

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So now let's discuss about the next step.

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So what we'll do, let's call let's try to find out the height and diameter of all subtree so left unsaid.

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So we'll call dysfunctioning diameter.

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We will pass the left necessary.

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So hydrometer function will give us the left answer, what is left unsaid, left and left, right and

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left DAYM and let's call the recursion.

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So this is right answer.

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Called the function radiometer and give it the right subtree.

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So what we are doing right now, so this is Salvatore.

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So I'm calling the function on the left subtree, it will return a pair of intelligent design, I'm

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calling it left unsaid, so it will return a pair left unsaid.

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So the first values are the second values Dimery, similarly, the first values site and the second

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one is damaged.

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So now we have left answer and we have a right answer.

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Let's find out these four values.

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So let's create variables.

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Integer left eight.

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This is.

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Left unsaid, darte first.

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So what is left diameter?

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This is.

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Left unsaid, dot second.

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Now, what is right, right, so this is.

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Right, answer that first.

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And what is right, diameter, so this is.

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Right answer, Nade second.

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So now we have all these values, we have these two values and we have these two values, and now it's

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time to find out the overall height and overall diameter and then we will return the pair.

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OK, so that is the function espere.

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So we construct a compared with the overall height and the overall diameter and then we will return

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dispair.

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So let's write the code for finding out the overall height.

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So in height, this is overall height and what is our own height, so this is one plus.

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Maximov left out a comma right out.

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And what is so little diameter?

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So.

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What is the overall diameter?

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So it has three options, so the maximum of what is the first option left out plus right eight second

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option is.

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Left VanMeter and third option is right to Dimitar.

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So we know what is all right and so we know what is right and what is our diameter.

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So now we need to return a here.

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There didn't have the function espere.

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So we will create a bear.

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So bear of indigenous or to be subpoenaed first.

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Is this all right?

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And be second is the overall diameter.

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So after constructing the pier, we need to return it also and return.

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B.

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So now I think this court will work.

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So let's run this court and then we'll submit.

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So now let's Amitava called.

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OK, so basically our goal is working.

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Now let's discuss the time, complexity, and then we will discuss how it will work.

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So first of all, let's discuss the time, complexity.

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So what we are doing here, we are calling it a collision on the left.

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We are calling it a collision, the right subtree.

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We are calling records for the left subtree, we are calling an occasion for the right subtree, so

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basically we are going to each and every node and let's see how much amount of work we are doing for

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each and so for each A. This is constant amount of work.

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This is constant work, this is constant work, this is constant work.

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So basically we are going to each and every node and for each and what we are doing constant work.

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So the total time complexity will be N.K..

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So the total time complexities we are going through each and every note and for each node, we are doing

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a constant amount of work.

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So engaged, basically same as big often.

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So now you can see in the previous solution, our time complexity was and multiply by eight.

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So now our time complexity is big off.

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And so this solution is much better than the previous one.

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And now let's take an example and now let's see how our solution will work.

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So first of all, let's take a very simple example, so if tree has only one or so, if there is only

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one node, but we will do so, we will call the function on the left.

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So for the left, Sabry.

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The height is zero and zero zero zero, you will call the function of the lights of the idea zero and

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the diameter is zero.

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Now for this node you will calculate the overall height and overall diameter.

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So what is height?

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Height is left height, maximum of left, right, right, eight plus one.

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So height is one.

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And what is diameter.

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So three option.

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So the first option is left out.

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Right out, which is zero level diameter, which is zero, right down to zero.

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So maximum of zero zero zero.

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So it will be zero.

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So basically for one node, if my tree has only one node, height is one and diameter zero diameter

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should be zero.

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Because if there's only one node, then diameter should be zero.

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Right.

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Because for this diameter maximum distance between two nodes, maximum distance between any two nodes,

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but there is only one node, so there will be zero.

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So we know if there is only one node.

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The height is one and the diameter is zero, so let's see, let's take one more example.

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So if this is Maitri.

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If this is tree, then what should be the output so high, it should be too and diameter should be two,

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OK, diameter should be two and should be two.

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So let's see how it will work.

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So for one node, we know height is one and diameter zero.

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Similarly, if there's only one node, we know height is one and diameter is zero.

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Now for this node.

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What is the overall height.

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So the maximum of left, right, right, right.

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Which is one and one plus one is two.

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So it is two.

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And what is diameter.

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So diameter is the maximum of left height plus right height.

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So which is two.

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Left diameter, which is zero right diameter, which is zero.

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So we to take the maximum of these three, so the output will be two.

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So diameter is two.

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So, Heidi, still dealing with those two and which is the perfect answer, so you will return the time

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you will get to commit to and you are returning the diameter.

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So now let us take a complex example.

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So let's take a victory as input and then we will see what we decide in diameter.

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So if this is my three, one, two, three, four, five, six, seven.

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Eight and nine.

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So what will be the correct height and what is the correct diameter, so height is basically one, two,

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three and four, so I should be fought for this tree and diameter is so one, two, three, four and

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five.

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So diameter should be five and it should be four.

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So when will on give me the height and diameter to call on for medium height and diameter and for one

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node for when or what will happen.

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You will go on the left, you will get zero zero.

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You will call right.

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You will get zero zero.

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And for one node, if there is only one node we know the answer will be one zero.

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Similarly, if there is only one note, we have seen that the height is one diameter of zero.

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Height is one dimwitted zero.

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Height is one diameter of zero.

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Similarly for this, Naude height is one diameter zero Lalaji.

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So if one or two, what is the height?

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What is the overall height?

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So height is basically a maximum of left and right, which is one plus one.

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So height is two and what is diameter.

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So diameter is basically left height plus one plus one is two level diameter is zero.

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Right diameter is zero zero zero and the maximum so two.

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So it will return to Clamato.

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Height is two, diameter is two for this node.

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So this is the same situation.

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So it will also return exactly the same thing, Heidi, is to Demetris to.

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So now let's see for node three to four, Naughtie, what is the height?

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So basically, it is Maximov, left, right, right, right, plus one, so for three, it will be three.

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And what is the diameter, so diameter 40 is basically left out, plus right out, two plus one is three,

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left diameter is two.

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Right, diameter is zero.

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That makes some of these two is three so high, it is also three and diameter is also three.

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So now let's find out the answer for one, two for one.

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What is the height?

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So height is maximum of left height and right eight plus one.

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So it will be three plus one four.

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And what is your diameter?

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So diameter four one is basically the left height plus right eight.

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So two plus three comma left diameter which is two right diameter which is three.

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So the maximum will be five.

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OK, so this value is five.

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So we need to take the maximum.

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So maximum is five.

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So diameter is five.

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OK, so what happened?

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OK, so this is Natalie, so now you can see.

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So now you can see the height is four and five, and which is the right answer, height is for and diameter

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is five, so you will get a pair here and you are returning the second value.

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So second value was diameter.

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So that's how everything was working.

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And the time, complexity of this code is basically big often.

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So this approach of returning to things from what we are doing so annoyed is calling on another note

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and that note is returning to things at once, this calculating both the things at once in only one

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location call, and it is returning to values, or you can say it is returning multiple values.

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So this technique, this approach is used in many problems.

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So, for example, if there is a problem in which you find out four things.

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If you were to find out four things at once, then you can use this approach.

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So what we will do so this class is only for two things.

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In this case, if you want four things at once, so you will create a glass, your own glass, let's

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say glass information.

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You want four things at once and you will write these four things first thing second, third and fourth,

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and then identify what the functional information so you will return an object of glass information.

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So this approach, this technique is very useful going forward.

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You may encounter such problems in that problems.

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In order to get a good time complexity, you need to return multiple things at once.

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And this is the technique for returning multiple things, for calculating and for returning multiple

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things at once.

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So this question is very important.

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So many companies ask this question, OK, so this question is very important.

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Diameter of a binary mining company ask this question in their interviews so you should know how everything

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is working.

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So this is it from this video.

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I will see you in the next one.