 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 mechanism or a technique that allows you to transform or change the data without adding or incurring any additional extra memory spaces. Let's go ahead and create a class, a Java class. We'll call it a Christian, and then the name of the class would be in place as I was the project demo. OK. OK, so first, let's go ahead and create the main method. I also want to create a function just to print information here. So we'll call it the static void print. I'll be using an array of just a single array, obj here, and we will print it out. So basically, for every object of n, obj, I'm going to print it out. It was shortcut SLUT, or you just use the print, the n, and then we'll put a white space after that as well. And then, yeah, that should be fine. And I also want to print a carriage return. OK, so that'll be the print function. I'm going to say just hide that for now. So first one, I'm going to do a, this is, I guess, I will call it in place using attempt, attempt variable. So just do a very simple swap. So first, let's create an integer list. I'm using the object down here for printing. So I'm going to create an integer using this integer class as opposed to the primitive type. We'll just call it n, and we'll just set some numbers here, like 10, 20, 30, 40, 50, but be fine. Just find the numbers. But you just imagine it's like a million in there, OK? So we're going to print it out first, print the n, the before, and print the after. So we know the before and after was a look like. OK, so this is the algorithm here. First, I'm going to loop through this entire list. The idea is that you're going to move right 5, 50 to the front, 10 to the back, and vice versa. So basically, as you move along here, you're going to not going to do all the way to the end. Basically, as you swap them, when you reach the midpoint, you can stop. So you don't have to do the rest of that. That's not necessary. That's when you are done. So we're going to do something like for every n i, this is the index position. When i is less than the midpoint of this length. All right, so in the length, then divided by 2, OK? When we reach the midpoint, we're done. All right, plus plus. And we just basically swap them out, right? So we create it in 10 first. This is the 10, which takes out just a very tiny space of memory, so it's not a big deal. So the space complexity here just, you know, it costs them really small. We'll replace that with the end of the i position, OK? And then we swap the i position, which is, if it's the first one, we're swapping this lot with the number from the end, OK? So you do n of the n-th length minus 1. And they also want to do it minus j. I mean, not j i, right? So that as we move along, as you move this one here, we're going to swap with this. If I move the next one, you're going to swap with this. So basically the length of this minus the i, so both sides will be moving all together, OK? Using this approach here. So that's what this minus 1 minus i here, OK? We swap that one. And then finally, we're going to swap the last position, which is n of length minus 1 minus i equal to the 10. So this is the classic swapping those numbers, OK? So I think that's pretty much it for this one. So let's see what's going on. Let's save it and run from here. And hopefully it works. It's going to take a while. So you can see down here, it's printing nicely. I'm going to move this to the right side so that it's top right top, OK? So here it is right here. If you change the size to A4, it should also work. And it run. I don't know if they're not. OK, so there it is. All right, so that is about it. So you can see here, we're doing it almost to get in place, almost 100%, but it's not. But in this case, this is a very tiny number, so it's not going to be a big deal, OK? So in the next video, we'll do a different operation. Similar one, but using a different technique. Thank you.