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Hi, everyone.

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So today we are going to sign the revised version of the problem that we saw in the earlier session,

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so cycle two.

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Now, in this problem, what we have to do, so instead of returning a boolean value, we have to return.

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The starting point of the loop.

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OK, so we have to return to here in the last problem.

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We have to return only to our fault.

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But in this problem, we have to return the starting point of the loop.

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So the starting point of the loop is to OK.

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So so this question is present on the record.

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OK, so for example, here, the starting point is to set out what should we do here?

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The starting point is one or at least the output is one.

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And here there is no cycle.

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So I will return null.

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OK, and our challenge is to solve this problem without using any extra space.

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OK.

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So this was our only problem.

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Now, how we can solve this problem, so the first approach of solving this problem is with the help

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of MAP.

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OK, we can solve this problem easily with the help of map, just like in the old session, just like

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in the earlier session, what we will do, we will take a map of Little North Star and value.

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So I will take a map of list North Star and a Boolean value.

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So what we will do.

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OK, so let's do this problem.

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So I have a map, eight percent inside the map.

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No one present, no more than nine percent inside the map.

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No.

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Then Bush for president.

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No.

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Then put forward to present.

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No, but to three percent or seven percent.

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No to president no.

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Nine percent.

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No.

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So put all of these.

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So three, seven, two and nine.

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Then again is to present inside the maps to is present.

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So in the only problem would be did the return to at this point.

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But here what we will do, we will return to OK, we have I will return my key value, I will simply

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return to the time.

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Complexity is basically when and the space complexity is on.

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OK, so let us implement.

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OK, so I closed the solution.

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So I will copy the same code just instead of returning the troops, instead of returning to enforce

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what they've done.

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We will them better than they normally do.

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OK, let's make this a.

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So instead of returning the true what they will return, I will return, OK, instead of returning false

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here, I will return.

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OK.

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So let's try to separate the so-called.

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OK, so our goal is working, so their time in the space complexities both our own.

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OK, so can we do better?

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Yes, obviously we can do better.

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So how we can solve this problem.

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So in the earlier session, we we discussed about the floor and the first point, that approach.

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OK, so how we can use this law enforcement approach.

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So if you remember the legislation both and slow and fast point that approach made at this point.

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OK, so both slow and fast point approach, they were meeting at this point.

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So after meeting at this point, what I will do, I will find the size of the loop.

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OK, so find the size of the loop to her finding the size of the loop.

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What what can I do here?

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So keep one pointer at three, for example, keep fast pointed at three and move the slow pointer.

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OK, so it's lowpoint that will move so slow will come here then slow will come here, then slow will

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come here, then snow will come here and then slow will again.

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Come here.

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OK, so while the slow pointer is moving around, what I will do, I will maintain account and I will

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do current placeless.

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So when the slope in the first point I will meet again, I will get to know the size of the loop.

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OK, so once we know the size of the loop, so let's say the size of the loop.

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

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So this is the loop will come out to be five.

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OK, so we'll say slope point will come here.

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So card will become one, then two, then three, then the will of current forward and then the value

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of count will become five.

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So we will get to know the size of the loop easily.

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Very easily.

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OK, so once we know the size of the law, what we can do, we will take two point.

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Again, we will take.

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Two points again, so I will keep the first pointer here.

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And what I will do with the second pointer to the second point, I will move five steps ahead.

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So step one, step two, step three, step forward and step five.

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So this will be my second pointer.

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And then both first and second point, they will move at the same speed.

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They will take one jump at a time.

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OK, they will take one jump at a time.

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So first comes here and second comes here.

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First comes here.

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Second comes here.

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First comes here.

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Second comes here.

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First comes here.

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And second also comes here.

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OK, both first and second will meet only at the starting point.

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OK, so both first and second pointer, they will meet only at the starting point.

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Why starting point.

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Because the difference between the second pointer and second pointer minus first point, the difference

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between them is basically key and what is fake is basically the size of the loop.

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OK, guys, basically the size of the loop, so the only point we had, the first and the second point

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I can meet is basically the starting point.

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They can they cannot meet anywhere else.

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OK, so with the help of this approach, our time will be open and our space will be open.

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OK, and this was the challenge.

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So in order to write the code.

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OK, so far, so what we will do.

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We will copy this code also.

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Slow and fast pointer approach.

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OK, so let's comment about this map approach.

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OK, so here we will not return anything, but we will do.

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We will only break this loop, OK?

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We will only break this loop and this means there is basically cycle cycle addicted.

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OK, so the cyclist predicted.

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Now, if there is not any cycle, if the cycle is not there, basically if the forest becomes null.

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OK, so if the fastest null light means cycle does not exist or we can see the next.

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The next of fast becomes then also cycle does not exist.

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So if this is the case, that means.

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Cycle does not exist, so if the cycle does not exist, we can simply eternal, OK, so there are two

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ways to come out of this loop.

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First is due to the big statement, that means the news cycle cycle is dead.

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Second is with the help of this condition, if this condition becomes false.

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So if the condition becomes false, that means there is no cycle.

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So I'm returning null, OK?

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Now, as discussed, so what we have to do, so I am summarizing the points, what we have to do, we

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have to first find the size of the loop.

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We have to first find the size of this cycle.

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OK, so we're finding the size.

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What we do, we will keep the first point where it is and we will move this little pointer around.

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And while this low point does meet is moving around, I will maintain the count, OK?

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Once we get the count, but we have to do we have to, again, take two pointers.

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Both the pointers will be present at the end of the list.

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So first pointer and the second pointer will move.

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Tekin Pointer count steps ahead.

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OK, so count will move.

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The second point, the count step ahead.

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Now the difference between the second and the first pointer is basically key now both the first and

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the second point that they will move at the same speed, OK, they will move at the same speed and the

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only point where they can intersect, where they can meet is the starting point of the linked list.

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OK, because the difference between them is gay and basically the size of the loop.

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OK, so I hope that makes sense to you.

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So now let us try to find the size of the loop.

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OK, so.

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Let's move slow point to one step ahead, so slow because the next Oslo.

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OK, and gay kwasman, OK?

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Now, while slow, is not close to fast.

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But we have to do so slow will move around slow equals next slow.

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And we will look, C++ is basically the number of nodes present inside the loop.

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OK, now at this point, we know the size of the loop, so I have to take two pointers here.

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OK.

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So I have to take.

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Two pointers, one and two, so they both will be present at the head of the linked list.

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OK, they're both fully present at the head of the list.

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The second point, the.

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Now, what we have to do is move the second point, 30 steps ahead.

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OK, so.

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Move.

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Second point steps ahead.

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OK, so while K minus minus.

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To is next to.

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OK, now what we have to do now move both the point that at the same speed.

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OK, so move both first and second pointer at the same speed.

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So move both pointer at the same speed.

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And they will only intersect at the starting point, so while one is not equal to two.

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So what we have to do seems beautiful when it goes the next of one.

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And two equals.

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Next of the.

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OK, so I think our gold is fine and it should work.

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OK, so once this loop is over, you can return one or you can return to, OK, I can return one or

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I can return to.

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OK, so our court is working.

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OK, so Al Gore is working now, the time and the space complexity of the solution is time is on and

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the space is big of one.

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OK, now there is easier way to solve this problem.

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OK, we have discussed we have discussed a solution for this problem.

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Now there exists a third solution, which is the most optimized solution.

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OK, now we will use the floor in the Fassbinder approach.

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So if we remember both slow and fast, Pointer will meet here.

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OK, so this lowpoint, those here and first point is here now after this low point in the first point,

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that approach.

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But we have to do.

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Move this little pointer at the head of the linked list again.

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OK, so remove the slow pointer from here.

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And put this low point guard here.

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OK, so the first step is move slow pointer to head.

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OK, now the second is what will do both slow and fast, move both slow and fast at the same speed?

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Move both slow and fast at the same speed.

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So what will happen?

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So after moving slow and fast, slow will come here.

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Fosterville, come here.

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Slow welcome here.

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Fosterville, come here.

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Slow welcome, come here.

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Fosterville, come here.

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Slow welcome here.

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And Fosterville, come here so you can see both slow and fast will meet at the starting point.

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OK, so what we have to do, we have to only take two steps.

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First, move the slow to the head.

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And then move slow and fast at the same speed, and they will intersect at the starting point of the

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loop.

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Now, why this is the scenario.

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So this is very simple mathematical problem.

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So first, let us write the code and then we will explain the mathematical solution.

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OK.

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So let us come out discord.

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So what we are doing here?

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So this is a simple solution, Fassbinder approach.

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OK, so once this low point and the first point I approach, what we have to do is move slow to the

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head.

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OK, so we have to do two steps.

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So the first step is move slow to the head and then move slow and fast at the same speed.

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OK.

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So move slow to the head, so slow it was a now slow is they're tired and they will only intersect.

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At this starting point, so while slow is not too fast.

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We have to move them, and this means we at the same speed, OK?

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Simple, and you can read Denslow or you can read them first.

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OK.

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OK, so our goal is working now, we have to write only this much code and we will get the starting

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point of the linked list.

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Now, how this is working.

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Let us try to understand.

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OK, so.

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So suppose.

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This is the cycle in the linked list.

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OK, and let's take some variables.

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So suppose this is this dense M and let's say this is Mr. Mansky.

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And so this is my intersection.

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This is basically the starting point of the linked list and this is the point where slow and fast intersect.

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OK, initially slow and fast intersect here.

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And let's say the size of the slope, the size of this loop is basically let's call it El, OK?

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Now, initially, so before meeting at this point, slow and fast pointer.

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So how many distance slow points will cover.

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So distance traveled by small point.

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That is basically this M.

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Distance plus, let's say the slope point, I will cover letters lowpoint, I will cover X around,

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OK, X complete round of the loop.

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So each round is of length L.

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So so this low point is covering the x rounds.

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Before intersection with the fast pointer.

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OK, so it's complete round and each round is of Lendell and then this distance key.

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OK, and what is the distance?

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So what is the distance travelled by the fast pointer?

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So distance traveled by a fast point that is first of all this distance M, so this distance M, let's

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say the number of rounds, the first point to go it ottavi.

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OK, so while each round is of length L and obviously this distance K OK.

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So what are X and why are you so X and Y are basically the rounds?

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Of the loop of the cycle before they intersect.

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OK.

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So so this is the clean bigram.

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OK, so this distance from the heart of the link to the starting point, this distances them.

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This is key.

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This distance is basically key.

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Advanced law enforcement intersect and the whole distance, the size of the loop is, well, OK.

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So I hope you know the equation.

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You know, both these equations will these equations are clear to you?

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OK, now one thing that we know, Fassbinder is moving with two x speed, OK?

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So if the speed of slow, pointillist X, then the speed of fast point is to X, that means the distance

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traveled by a fast pointer is twice the distance travelled by this little pointer.

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OK, so in mathematical terms, so distance travelled by a fast pointer is twice the distance traveled

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by the slow pointer.

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OK, so let's put it all here.

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So this is M plus right into L Klasky.

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OK, so twice and distance David Westlaw point.

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That is M plus X in into L plus K.

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OK, so what will happen.

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And plus Y into helpless K and multiply to inside so to M plus two excel plus 2k.

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OK.

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Now, what will happen, so take em decide, take the side and take to excel, decide, OK, so what

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will happen?

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This will become V minus two.

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And 2L is basically.

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McCleskey OK, so let me reiterate it.

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So basically.

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McCleskey.

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Is why minus two weeks and total.

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OK, and this is our simple diagram.

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So this is M and this is K.

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OK, so this is.

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Now, see, one thing here, emplace gay is an integer, multiple of L.

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What is L.

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L is basically the size of the loop.

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OK, so McCluskey is in danger, multiple of.

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Integer, multiple of L.

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OK, now can we put one here it is integer multiple.

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OK, so why am minus two weeks will be integer.

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OK, so it can be anything.

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Now suppose it is one.

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Suppose it is one.

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So basically what I'm writing here K is basically an OK, so suppose this value is one.

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This value can be to, this can be three, this value can be anything.

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But for the sake of understanding let's suppose this value is one.

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OK, so emplace is basically L.

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OK, so if this, so this distance is key, that means this distance is M ok because the total distance

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is L.

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OK, so the total size the loop is.

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And so if this much distance is key, if this much distance is K then obviously this will be M.

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OK, this is them.

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Now see here, so slow and fast pointer, slow and fast.

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Both are present at key.

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So I am drawing this diagram once again.

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This is both slow and fast pointer are presented.

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OK, so this distance is key.

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This distance is M and similarly, this distance is M.

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OK, so both slow and fast pointer are present here.

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Now what we did, we moved this slow pointer.

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To the head of the linked list, so this is Logan.

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Now, what we are doing, we are moving slow and fast at the same speed.

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OK, slow and fast are moving at the same speed.

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Now, what will happen when slow will cover em distance slow will reach here, slow will reach here

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after conveying the distance.

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And at the same time, fast will also cover the distance.

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M fast will cover the distance and it will also reach here.

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And that's how the code will work.

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OK, so I hope you understood the code.

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OK, so that's how this is working.

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If you have any doubt, feel free to ask.

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Thank you.