 Hi, this is Christian. So in this video, we're going to create an algorithm to perform what's called an in-place operation. So an in-place algorithm is a technique to transform or change the data without adding or incurring any additional extra memory spaces. And this time, we'll use another algorithm called just basically simple math addition subtraction stuff. So let's create this first. We'll call it in-place add subtract, OK? And integer, and so, OK, for this operation to work, basically, it's quite simple. And this is another, I guess you can say, pure in-place operation. How does it work? If you want to swap the 10 and 50, OK? How do you do that? If you think about this, if I replace this position here by adding 50 to 10, if I do, let's say, n of 0, it's equal to n of 0 plus n of the last one will be 4, right? So basically, if I do that, I get a 60, OK? This is 60. Good. So now, how do I change that one? So I'm going to go n of the last one. In this case, it's 4. It's equal to the n of 0 subtract itself, so n of 4. If I do that, what do I get? I got a 10, OK? So now, 10 is now replaced right here. So this is now replaced with a 10. And this is now replaced with a 60 already. So how do I get that 50 back over here? Well, same thing, right? Similar, you get n of 0. It's equal to the n of 0 subtract n of the fourth position. I get my 50 back. I just swap my 50 and my 10 without incurring any additional memory space, OK? So this is another really nice example. You could do subtract first, and then you can add it, and then subtract it back. It's just a really simple math operation. All right, so for every n i less than the n that length by 2, and then i plus plus, OK? What I did here is basically divided by 2, OK? Because again, just to make sure you understand why, I just want to do halfway, right? Once you do halfway, you already swap everything. You don't have to go through all the way of the array. That's not necessary. So now, what I do is I'm going to get the n of the 0 position, which is i, equal to n of i plus n of the last position. That's the n of the length of 1, OK? Minus 1 minus i, right? You plus that. And now that's 60, OK? And then n of the last position, length minus 1 minus i, is equal to the n of the first position, which is i minus its current position, n of the length minus 1 minus i. And we get our n number of that, which is now a 10, OK? And then finally, we do it the other way around. It is now equal to the n of i, subtract the n of the n that length minus 1 minus i, OK? I think that's it. So let's give it a try and see if this works. So let's go back up here and call our function and place add subtract, pass in m, OK? So say this, run, and there it is, right? 1, 10, 20, and then 5, 15, back and forth.