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Hi, everyone.

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So in this video, we are going to write the court for this problem and let's start writing the code.

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I have already explained you everything.

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So now the only thing remaining is to just start writing the code.

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So, first of all, we should change the name to in order.

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Right.

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We are given in order areas input.

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And let's discuss the recursive relation once again and let's write it somewhere.

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So we discuss that.

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First of all, we will find out the maximum element in the area and obviously the index for this and

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let's call it mixed index.

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We will find out the maximum element and the mix index.

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So what we'll do after finding out the maximum element, this will be my root node.

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This will be my node for calling on left.

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I will pass this index and I will pass.

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Max index minus one.

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So this is the START index, this is the end index and for calling on right, the START index will be

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mixed index plus one and an index and that's it.

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And that's third rate.

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So let's start writing the code.

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So first of all, let's check the size.

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So if the size of the area and the size of zero, that means the air is empty.

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So in this case, your tree will not exist so you can return null.

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Otherwise, what I'm going to do is write a function helper.

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So what this helper function does, it creates three.

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And I need to pass the in order that I will pass the starting, which is zero starting index and I will

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pass the ADD index, which is in order that says minus one.

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Fine, this is the START index and this is the end index and this helper function, it will return the

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routine order of the Nutri.

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So what this helper function will return a dilettante root node.

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So it is a train or start helper function, what it is taking as input, it is taking in other areas

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input so you can take in order that is in port.

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What else.

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So it is taking the starting next.

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It is taking and index as input.

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So this is going to be a recursive function.

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So we know the recursive relation, but what will be the best case?

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So basically it is very simple.

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If START is good, then this is a start.

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So if start is good, then end.

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It started as guerdon, and that means they're done.

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We will return null now.

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What we need to do, so what was the first step?

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First step was basically we need to find out the maximum element, because that will be my route to

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find a maksym element, because that element is my route.

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So let's take one variable, let's say the maximum value, so what is the maximum value?

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This is in the minimum and obviously we need maximum indexed so that I can call on left and right.

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So let's take one more variable, which is maximum depth.

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And let's say initially this is minus one.

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Now we're trying to do.

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We need to iterate over the area to find out the maximum element.

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So I will start from starting index I list in articles two and index I placeless.

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So if the current element.

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Is greater than the maximum value, then we are going to update the maximum value to this value and

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obviously we will update the maximum index.

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So Maksym indexes, I find.

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So now we know the maxim element so we can create root node, so let's create a root for the Soul Train

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Wildstar Root will be new right node and we need to give the data that is the maximum value.

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Fine.

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So now for this route.

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And to find out the left subtree for this and we need to find out the rights of this and after finding

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out these two values, we will return.

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So what will be left?

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We discussed that.

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What I need to do, I will call on left separately.

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So in order, it will remain same.

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Starting next will remain the same and what will be the end and will be the maximum index minus one,

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and for building the rights of Prevert, I will do the start.

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Index value will be the max index plus one and and index and the two guys that said, so what we are

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doing here, this is simple case, right?

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So our approach was, first of all, we need to find out the maximum element late because the traversal

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is left-to-right, the maximum element will be our root node.

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And obviously we need to find out the MAKSYM index so that I can call on left and right.

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So I am migrating from start to end.

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And if this data is greater than the maximum value, then you will update the maximum value and the

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maximum index, which is simple right now.

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What they are doing, since we know the maximum data, let's say the maximum value is five.

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So I'm creating a node with the data five.

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So we will create an order of data five and four, building the three.

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We need to build the left of the tree.

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We need to build right on the tree.

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So I am building left and building right for building left called the helper function for building right

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called the helper function and the end will be maksym index minus one and the start will be in this

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case maksym index plus one.

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And that's a very simple code.

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So let's try to run our code first.

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We will test against the basic test cases, then we will submit.

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OK, so the spelling mistake, this will not be try this is actually three.

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Yep, so let's admit now.

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So our goal is working, right?

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This was very simple called.

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Only this much amount of gold, so this was really, really very simple.

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So this is all about this video, guys, I will see you in the next one.

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Thank you.