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Hi, everyone.

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So in this video, we are going to discuss and find the diameter of a binary tree.

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So what is the meaning of diameter?

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So encircle?

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Diameter is basically the maximum distance between any two points, so similarly, in case of binary,

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diameter is the maximum distance between any two nodes.

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So let us take one example.

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So this is a binary tree and what is the distance between this node and this node?

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So count Edwyn as two and three.

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So the distance is basically three.

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And this note and this note, the circumstances, one, two, three and four, so the distances for

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but I need to find out the diameter and diameter is the maximum distance between any two roads.

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So the maximum distance will be between this node and this additional.

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So count one, two, three, four and five.

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So the answer is five.

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Now, there are many ways in which we can obtain five.

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For example, the distance within this node and this node is also five.

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The distance between this node and this node is also five.

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So there are many ways to achieve this number five.

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But since we only want to return the integer so we do not care about the number of combinations, we

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just want to return the diameter and diameters, the maximum distance between any two node.

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So for Destry.

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My output will be five.

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So how we can solve this question, so let's take one more example first.

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So if this is a binary tree.

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What is the maximum distance for this diameter?

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So one, two, three and four, so the diameter is four and we can see we can say with this diameter,

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diameter is left eight plus right height.

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So left out is two.

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And right now it is also two.

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So two plus two is four.

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So let's take one more example.

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So for this binary tree vertebrae, the diameter.

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So diameter is between this node and this node.

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So one, two, three, four and five.

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So diameter is coming out to be five.

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And what is left, right, so left out is to what is right.

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So right now it is three.

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So two plus two is five.

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So can we say that this formula is correct?

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So I think this is a good guess, but this formula is not correct.

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Why?

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It is not correct.

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So I will give you one example.

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This formula will not always work.

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It will work only in some situations.

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So if this is Maitri.

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So for this Pinetree.

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What is left out, right, right, right, right is one.

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And left is one, two, three and four, so four plus one is five, so according to this formula, my

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diameter should be five.

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But let's see whether that's correct or not.

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So I think it is not distance between this note and this note.

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So one, two, three, four, five and six.

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So the correct answer is basically this.

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So in this problem, in case of this binary tree, this one is not correct.

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So this is a very good guess.

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But this is not correct.

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This is not the correct way of finding the diameter.

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So in this case, the diameter is not passing through, the diameter is not passing through the route.

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And I never said in the question that diameter must pass through root.

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So it is not necessary that diameter must pass through root.

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So what are the situations that we can have?

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So there are three situations, according to me.

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So first one is basically how we will utilize our tree so we will regulate our tree like this.

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Now, one situation is the first note lies in the left, the second northerlies in the right supply.

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So in this case, our answer is correct.

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So diameter will be left eight plus right out.

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Now, if both the nodes lie in the left Sabry so if both an old line, the leaves are so right it doesn't

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have any effect.

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So in this case my answer will be left diameter.

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So left diameter is the diameter of the whole tree, so diameter will be the diameter of the hole.

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So this is whole the diameter because both the nodes lying in the left and right subtree has no effect.

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Now if both the nodes are present in the right subtree, so in this case, my answer will be the right

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diameter because diameter does not have any effect, because the left tree does not have any effect.

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So in this case, the right diameter will be the whole tree diameter.

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There are two situations.

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So first one is basically our tree will be like this.

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So if this is Maitri.

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Then the second situation, military can be like this.

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And in the third situation, my dream can be like this.

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So for this kind of free for distri, my answer is basically left out, plus right out.

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So basically this thing.

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Left, right, right, right.

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So this will be the time for this kind of tree, my answer will be the left diameter monster will be

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the diameter for this tree.

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My answer will be the right diameter.

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So what we need to do.

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So we'll find out all this value.

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So this is our first option, this is our second option and this is our third option and our answer

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will be the maximum of the three options.

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So make them our first option, second option and third option, simple and how we can find out the

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diameter we will colocation for finding out the diameter.

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Similarly, we will call the for finding out the right diameter.

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And that will be my answer.

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Simple.

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So this question is basically present only it could find the diameter of a binary tree and now let us

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compute the cord for this function.

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So we are taking root as input.

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First of all, we need to discuss about the best case, so what is the best case if the route is.

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So if rotational, what is their diameter, diameter will be zero, simple maximum distance between

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any two nodes and in my tree I don't have any node, so diameter will be zero.

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So our first option is basically.

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This one, left, right, plus retied.

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So left out, right, right.

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So I will write the code for this hate function.

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I will give the left subtree.

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And I will give.

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The right subtree.

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So I will write the code for this function, how to function.

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So our first option is very simple.

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Left, right plus right now our second option is.

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I was separate option is left Daym and third option is right diameter.

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So our option to.

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It's basically Dimitrov by Nutri.

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And we will give it left subtree.

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And our third option is.

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The right diameter, so we'll call this function.

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And we will give it.

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The right subtree.

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And now what we need to do.

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So we need to lay the Maximov first option, second option and third option, so we have inbuilt max

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function, so we will use it.

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So Maximov.

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N.B. This is how it works.

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It will take two things as it put to argument as input.

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And I want to take the input of two things.

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So I will write like this mix of a mix of BNC.

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So this is how we can take the maximum of three things.

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So let's write the code.

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So I will return.

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The maximum of.

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Option one.

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Mix of option two and three, option two and option three.

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So the only thing left is basically to write the code for this function, so let's write the code.

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Will teacher, then the function is height.

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So this will take root and put.

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And redecide, so if.

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Rotational.

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My height is zero, if true, does not exist Diatta zero, otherwise the height will be one plus maximum

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of left and right out.

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One plus maximum of left out.

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And right, right.

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So, yes, I think our gold will rock solid right now, and then we will submit.

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Now, there are some rhetorical.

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So basically, our goal is working and now we will discuss the time, complexity for our solution,

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relative time, complexity.

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So first, let us discuss about the time, complexity of the function.

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So let's discuss about the to function.

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So what we are doing.

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We are calling it a collision on the left and on the right separately, so this is Natalie.

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So I will call it function for the left subtree, I will call it function for the right.

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So basically we are going to each and every node and for each node we are doing a constant amount of

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work.

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So for each node we are doing.

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So this constant amount of work, one plus.

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So this is also the constant amount of work.

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So we are going through each and every node and for each and what we are doing, constant amount of

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work.

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So that time complexity for how it function is big often.

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Now, let's try to find out the time, complexity of this function diameter function.

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So, again, what we are doing.

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So this is Maitri.

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So we are calling their diameter function on the left.

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We are calling diameter function on the right.

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So I am calling diameter a function for the left subtree.

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I am calling diameter function for the right subtree.

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So we are going to each and every node and let's see at each and how much amount of work we are doing.

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So this is constraint, the amount of work and then we are calling the function for every node, we

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are calling the function for every node and this is the constant amount of work.

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So basically for each node, I'm calling the function.

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So now let's try to find out the complexity of this function.

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So basically three can be of two types.

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So first three can be balanced.

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So this is an example of a balanced tree.

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The number of nodes in the left and the right are they both are the same.

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So I'm standing here at the root node and then what I'm doing.

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So for each node, I'm calling the function.

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So if there are any node, so for each and what I am calling the function so.

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Gay and minus one, or we can simply write it again.

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Plus, we are doing some constant amount of work for each a..

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OK, so this constant so we can remove it and then what we are doing, so we are calling the function.

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For delectably and for the rights Sabry, so two times.

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And by do OK, now we have seen this, so we have seen this somewhere, so this is exactly like.

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So this is.

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Exactly like Mozart.

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So what is your time, complexity, so time complexity is and Logan.

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Now, another kind of tree that we can have is basically this kind of tree.

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Now, let's try to find out the complexity of this function for this type of.

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So initially we are standing here at the root node, then I am calling.

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So this is constrained amount of work.

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So let's try to their number of nodes is.

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And so this constant amount of work or we can remove it since it is constant, then we are calling the

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hate function.

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So left simply does not exist and the right subtree.

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So how many number of nodes are present and minus one node.

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So I am Complexo and minus one time complexities and minus one in tookey simple Hyett function time.

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Complexity was linear if you remember so and minus K and then what we are doing.

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So we are calling the function on the left subtree.

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We are calling the function the right subtree.

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So left are Presnell and the rights are pre how many notes are there and minus one nodes are there.

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So this is the Rickerson deletion.

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So this is direct resolution.

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Now, what would be the time, complexity, so how we can solve this?

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So let's take the example of the bubblehead.

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So in the world sort.

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What we do, so we compare the elements and we still have them, then we compared the elements and we

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swab, we compared the element and we'll swab.

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So after one iteration.

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The largest element will represent at last.

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So after midnight, that means after End-Time big oftentime, the largest element will be present at

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last.

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So we have to do this same thing for and minus one element we have done once.

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And it cost me big oftentime.

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I have to do the same thing for the remaining and minus one element.

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So what is the solution?

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So for one element, you did big oftentime.

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And we need to do the same thing for nd minus one element also so and minus one.

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So now you can compare so and minus one it took, it seems.

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And so the reconciliation is the same.

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And what is your time complexity for the short term?

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Complexity is and square.

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So for this type of tree, the time complexity is and square.

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So let's write everything again.

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So we can have two kinds of free.

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So this is the first one, which is the balanced number of free number of nodes in the left, Sabry

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is equal to the number of nodes and the right subtree.

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And this is another kind of reach that we can have.

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So for Destry, the time, complexities and log-in, for distri, the time, complexities and Scribd.

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So what is the correct term complexity, so our role is we will report the worst case time complexity.

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So ideally we should say the time, complexities and square, because this is the worst case, time,

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complexity.

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But we will not write our answers and square because if we see for this kind of three.

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We can do the calculation and we can find out the height is basically log often where any number of

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nodes log often.

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And for this tree, there is no need to do any calculation for finding out the height.

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Height is simply the number of nodes present.

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So what is the time, complexity?

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Time is basically in multiply the height, Logan is basically the height for this type of reform.

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Balestra.

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So here the time complexities, number of nodes multiply by eight and here the time complexities and

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multiply by height and how it is and.

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So my correct term complexity is basically big off number of nodes multiplied by hate.

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So this is the correct time complexity that we will report number of nodes multiplied by HAID.

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So what we want to do so in the next we do, we want to remove this, we want to decrease our time,

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complexity, and by the time complexity is high, let's try to analyze why the time complexity for this

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function is high.

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So let's take one example.

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So if this is Maitri.

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One, two, three and four.

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So I am standing at this node, so what I will do, I will say find out for the left does not exist.

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So finally, hired for the right one will call on to give me the right.

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So I want to find the out of when no one will ask her to give me the hate.

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So we will find out ASAP and it will return the answer.

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Simple.

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So now option two so left does not exist.

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So option three daym so haven, give me your diameter.

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So when will call on two and in two we are again we will again call the hate function so four two two

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will ask for it again.

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So see when I am calling this function for two then two will call hate again.

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We will again ask for hate.

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So you can see we are doing repetition of the same work.

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We are doing the same work multiple times.

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So number of calls for two will increase, similarly, there will be more calls for this note, there

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will be much more call for this.

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So the time complexity is big off and multiply by eight by the time complexity is more because we are

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doing the same work many times.

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And that is the reason, so how we can improve our time complexity.

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So what we will do?

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If this is.

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Mitry So I will ask her to give me the heightened diameter at once.

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So to Willetton, height and diameter at once.

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OK, we will not call multiple time, so here what we are doing, first of all, I am asking so if this

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is the three.

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So for one, for one, I want to find out the hate for when I find out the hate, so I want you to give

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me the hate.

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So when will call on two for finding out the hate?

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So when I am the function of two, so two will again call on three hate, three give mediate or two

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will call on three to give it tight.

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Now, when you want to find out the diameter of one, so you want to find out the diameter of one,

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so one will call on to here to give given a diameter, we will again call for finding out the height.

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So we will again.

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So we are again finding out the height and for finding out who will again call on three and three will

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return the height.

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So we are doing the same work multiple times and that's why the time complexity is more so.

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What I want to do is I'm giving you a hint.

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So what I will do so in this approach, first we are asking here to give me the height.

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So once to give the height, then I will ask again here to give me the diameter.

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First I am asking for to give me the height.

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Then I'm asking to again to give me the diameter.

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OK, so one is calling on two.

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Here to give me the height, so went to Willetton, the height, then I am again calling on to find

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the diameter.

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So that's why the repetition of work will be there and then to Willetton, the diameter.

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So what I want to do is I will call in such a way.

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So I will call like this here to give me the height and diameter at once.

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So too will do the calculation.

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Only one, only once.

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And to Willerton both values and height and diameter both.

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So that's how we'll be able to solve this question and less time complexity so correctly, our time

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complexities and multiply by each.

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So we want to remove we want to decrease our time complexity.

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And this is the way that we will reduce our time complexity.

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So in the next we do we will discuss about it and we will also write the code.

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So this is it from this video.

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I will see you in the next one.