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Hi, everyone.

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So in today's video, we are going to solve this question and find the midpoint of a long list.

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OK, so one link left.

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Let's say the linked list is of Aadland.

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So one, two and three, four and five.

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OK, so what is the midpoint?

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So my point is basically this node three.

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OK, so what?

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We have to return.

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We have to return.

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The return of the function will be node star.

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And let's say the name of the function is find out what it will take.

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So it will take had input.

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OK, so we have to write the code.

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In this case when the Linklaters of all and then three, the middle node.

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So I have to return the node three right now.

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Let's consider this case when the length is basically even so one, two, three, four, five and six.

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So in this case basically the length is even so there are two, three and four.

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So two me are there now out of these two.

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I have to return the second.

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OK, so I have to return the second node.

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So I have to return for.

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So I hope you understood the question.

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Now let's see how we can solve this question.

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So first way of solving this question is basically first find out the length of the linked list.

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OK, so in this case, what is the length of the list five?

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In this case, what is the length of the six?

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OK, so first step is done now in the second step, what we will do.

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So you will do length by two, you will do length by two.

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So five by two is basically integer, one integer is basically two.

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And similarly what is six by two is three.

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Right.

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So this is index zero, index one, index two and three and index four.

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Now the third step is basically iterate how many times leant upon two times.

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So you can see lent by two is two and you can see.

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So the node three is present at index also first and second iteration.

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So you have to add rate length by two number of times.

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Simple, right.

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And in this case also lent about two was three and you can say zero.

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You can see zero, one, two and three.

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So the node for its present at the the N-word is index three.

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So index three is basically linked by two.

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So we have divided three times one time, a second time and basically third time.

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Right.

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So we can solve this question by finding out the length.

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Then we will do length by two and then we will get that much number of times simple now.

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So basically we are traversing the linked list two times, so we are traversing the linked list two

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times.

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So first time we will traverse to find out the length.

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Right.

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And the second time we will traverse the linked list to find out the middle element and we will do length

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by two.

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OK, we related leant upon two times.

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So if I will use this approach, finding the length first and then finding out the middle element.

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So I am traversing the list two times.

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OK, so I hope you understood what is the meaning of two times or two times.

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What I want to say is first of all, you have to find the length, right?

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First of all, you have to find the length.

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That means you will I draw the link list.

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So this is one traversal now, the second time you will like trade over the link list and length by

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two number of times.

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So this is your second traversal.

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So that is the meaning of two traversal.

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OK, so this is we are travelling two times.

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But the question is, let's say the question is find the midpoint of the linked list in one traversal

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in only a single one traversal.

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OK, we cannot traverse the list two times.

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We have to traverse the link list only once and then find out the better part of the linked list.

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So can you solve this question?

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So it's very easy.

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So this approach and the approach is called slow and fast pointer approach, slow and fast pointer approach.

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Now, what this approach says is basically so consider or lengthening list one, two, three, four

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and five.

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OK, and we will also consider even lengthening list one, then two, three, four, five and six.

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So what is slow and fast pointer approach?

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So it is a very simple approach.

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It is a slow pointer at the hard and fast pointer next offered OK and similarly slow pointer initially

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slow pointer will be present at the hard and fast pointer will be presented next of.

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OK, so first step is basically initialization.

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Now this approach is slow pointer will make will take a jump of one and fast pointer will take two jumps.

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OK, so what is the meaning.

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So I want to say slow will take a jump of one.

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So slow equals next of slow and fast will take a jump of two.

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So fast equals next of next of fast.

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OK, so the logic is slow.

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Will take a jump of one correct and fast will take a jump of two.

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So this is taking a jump of one and this is taking a jump of two.

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So the logic is so finally we have Nele here and finally we have Nele here.

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So the logic, the main idea behind this approach is basically when fast reaches none.

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OK, so when Fosterville Regional, then basically Uslu will be present at mid.

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OK, understood.

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So this is the core idea of this approach.

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So what is the idea?

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So you have to point us slow and fast.

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Fast is taking a jump off too slow is taking a jump of one then basically when.

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Fosterville reach the end of the limitlessness only means the end of the list, then basically global

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reach at the middle of the list, right.

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It is very obvious, right?

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Fastest moving twice the speed of slow.

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OK, so the fast speed is twice the speed of a slow.

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So fast is moving with quite the speed of slow, two times the speed of slow.

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So it is obvious when Fosterville reach the end of the linked list.

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This is the end of the list.

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Then slow pointer will reach the middle of the list.

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So let's see how.

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OK, so Slovik, take a jump of one.

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Oslo is here, and it simultaneously forced you to jump off to so fast, takes a jump off to.

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Now, again, snow takes a jump of one, so snow takes a jump of one, and Fosterville take a jump of

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two so fast takes a jump of two.

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Now you can see when fast reaches nul, then very slow.

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So your slow is present at the midpoint of the enclosed right slow, especially at the midpoint of the

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lift.

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Now let's consider even case also.

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OK, let's consider even case so slow takes a jump of one.

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So this is your slow and fast will take a jump of two at the same time.

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OK, so this is at the same time simultaneously and fast jump so fast we'll reach here.

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Simple.

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Now a slow takes a jump of one.

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OK, so this is my slow and fast take the jump off to.

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OK, so this is your first.

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Now, one thing to observe here is can fast take two jumps now.

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No, it is not possible.

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It is not possible because the next fastest.

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OK, so here we cannot take two jumps and why we cannot take two jumps because fast or next.

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So this part is null.

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And I told you what is the rule.

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So whenever you have something and after that you have next.

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That means this thing should exist.

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And if this thing does not exist, then basically you will get a random error.

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OK, so this thing cannot be this thing cannot win an otherwise you will get married.

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Now if you will see carefully.

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So the next of fast is basically nil.

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So I can not take two jumps.

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That means I have to stop here.

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I have to stop here.

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So if I'm stopping here, check slow, slow, slow is at three and three is not actually the end for

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both Halmet.

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But I have discussed above that we have to return.

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OK, we have to return second made and not the first bit.

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So what we will return.

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We will return slower or next.

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Simple, right.

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So what is slow.

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Slow is basically this law and what is slower next.

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And this is basically this not OK, simple.

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So that's how it is working now.

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Final rule, OK.

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So one last time discussion.

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So in case of old it is very simple then fast in case of fast will become fast will become null and

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you will simply return slow.

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It will simply slow.

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You can see so this fast, this fast actually becomes null.

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And this is your solution which is at the midpoint.

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Now in case of even it is a little bit different, fast will not become null.

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In fact, the next of fast is becoming well.

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OK, you can see.

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So this is your first.

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So the next hour fast is becoming null.

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And that case you can do Denslow, but in the end it is saying that you have to return the second and

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not the first.

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OK, so for returning the second limit, I have to return slower next.

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OK, simple.

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Fine, and initially we know initially, so this will be my slow and fast will be the slow will be ahead

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and fast will be headed next.

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OK, so this is next.

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And I hope you understood the question.

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Now, this requires only one traversal.

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So let me write the call and then I will show you.

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OK, so this question is basically present.

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Don't let go.

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OK, so this question is present only today and we will write the code here.

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So we have the written list not start.

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OK, so the name of the function is Matalote and we are receiving as input.

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And if you read the question, the question is, so if there are two middle nodes, then you have to

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return the second model.

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OK, fine.

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So as discussed.

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So what we discussed.

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So I told you, if you have a long list to one, two, three and five, so I'm taking two pointers initially,

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this will be my solo and this will be my first.

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And in case of forward, what will happen so your fast becomes null and you will Denselow.

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In case of even the next of fast becomes null and according to the question, you have to return.

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The next off slow, OK, so slow arrow next you can see so you can we have to return the second model

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symbol, right?

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So now let us write the code.

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So let's stick to Pointer North Star, slow, slow is presented, OK, initialization and very fast

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present.

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So fast is president at the next office.

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OK, so you can see slow is initially here and fast is here.

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So now we have to do so.

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We have to hydrate.

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So while fast is not to cause trouble, but you have to also.

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Right the next hour, fast is also not because fast is going to make two jumps.

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OK, so if this condition is too slow, takes a jump of one, so slow equals next slow and fast takes

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a jump of two.

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So fast equals.

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Next of next hour, fast, simple, right, when you will come out of this very loop.

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So there are two ways of coming out of this loop.

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Either the fast will become known or the fast growing next will become known.

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OK, so if basically.

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If fast around external.

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Or you can say so if fast exist.

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OK, let me write like this.

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So if faster than external, that means

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so that means basically you have equivalent, right.

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So if you have even learned what you will return, it will return slower next.

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Otherwise, what you will have done, you will return slow in case of.

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OK.

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So this is in case of hard.

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So I think this is pretty much that we have to do.

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So let's see how it is working.

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So this is I told you, this is the initial step again, the initialization step.

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OK, fine.

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And see, I told you, whenever you are putting next in ahead of something, then basically that thing

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should exist right.

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So fast.

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Next should exist.

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So that's why I wrote this condition.

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OK, and should exist because you are putting next in front of us.

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So that's why this condition and obviously we have to choose and operator simple right now, when you

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will come out of this, why loop.

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So either you have our lengthening list or do you have event linked list.

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So if you have even lengthening list, then basically this is the condition.

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So I am checking if the next of fastest and then yes, it is then basically as decided we have to return

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this next.

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So we are returning the extra flow and in case of Ford, you will simply return slow.

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So that's why I'm returning slow in case of Aadland linked list.

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So you can see here this is only one traversal.

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So we are traversing the linked list only once.

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We are traversing the linked list only once.

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OK, so we are doing single traversal to find out the middle order in our previous approach.

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First we were finding the length and for finding the length of one traversal and for finding the middle

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road we have to do we have to iterate at length about two times to find out the middle road.

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OK, so there was two traversal.

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Now this is one traversal.

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So which one is better?

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This approach, this slow and fast pointer approach, so slow and fast pointer approach is very good.

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OK, so this is better than this approach.

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All the time complexities seem right.

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So time complexity in both cases that big often.

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But in this approach we have one traversal.

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In this approach we have to traversal and law enforcement approach is very important and it is used

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in many places.

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So the logic was very simple.

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I am repeating myself fast is moving two times the speed of slow.

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So when fast will reach the end of the list, then slow will basically reach the middle of the list.

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And that is the logic behind this approach.

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So now let us test our solution with our solution is basically correct or not.

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So let me test our solution.

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So one thing that we did wrong here is basically we are using Nolde and you will see the structure.

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So we have listner, OK, and we have to return restored and not in order.

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So let's replace this.

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This is not the civil list.

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OK, so this will be listner and I think so this is held right.

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OK.

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OK, yeah.

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So first we will run this code and then we will submit.

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OK, so first run code.

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So we are getting an error here, OK?

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So can you guess why we are getting an error here?

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So it's very simple.

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I did a very silly mistake.

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So what we are doing here, we are using we are writing fast eru next.

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OK, so at this time we are writing faster.

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Now what if fast isn't a.

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OK, so what the fastest then we are, we will get a right.

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So instead of taking fast next, we're going to begin simply check if fast exist, we can simply check

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it first exist.

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So let me do that.

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So instead of writing this condition, you can simply write it first exist, so if that means this is

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even and I can assure you see in this case, in case of.

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So in case of Ford fast becomes.

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That means in case of even fast will not be.

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OK, so in case of even fast is not only.

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Simple, right?

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So I'm using this condition in case of even faster, not so in case of even fast exist how to use this

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line if fast exist or you can simply ride like this.

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So if fast is not questionable, voters seem OK now.

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Rancourt again.

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OK, so accepted.

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So this was the input provided by the this website and I would expect it out.

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This is our output three, four, five, and this is the expected output, three, four, five.

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And how this is the expected output, because what it is doing here is we are returning the middle order.

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So we turning three and this in the back end of late.

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So what they are doing, they are printing everything starting from the middle to the end.

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So they are printing three, four, five.

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And the expected output was also three, four, five.

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OK, so that's how it is working.

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So after I accepted, now we will submit our code and let's see whether our code is correct or not.

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So let's meet our code.

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So here you will see the solution.

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OK, so success means our goal is working and our goal is basically faster than a hundred percent of

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the C++ submission made for this question list.

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OK, so I will share this quote with you.

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So if you have any doubt in this question, you can ask me.

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OK, thank you.