In this video, we will study one of the simplest sort of algorithms called selection sort. So it's main idea is quite simple, we just keep growing the sorted part of our rate. So let me illustrate it on a toy example, assume we're given a sequence of links. Five consistent of five integers, eight four two five and two. So we start just by finding one of the minimum elements in this array, in this case it is two. Now lets just do the following, lets just swap it with the first element of our array. After swapping, two stays in its final position, so two is the minimum value of our array and it is already in its first position. Now let's do the fun one, let's just forget about this element. It is already in its final position and let's repeat the same procedure was the remaining part of our array. Namely, we began first find the minimum value, it is again two. We'll swap it with the first element of the remaining part and then we'll just forget about this element. So again, we find the minimum value which is now four with what was the first element of the remaining part which is now the sole element of our array. And then, we just forget about first three elements and we continue with only remaining parts. So once again, we just keep growing the sorted part of our array. In the end, what we have, is that the whole array is sorted. The pseudocode shown here on the slide, directly implements the idea of the selection sort algorithm that we just discussed. So here we have a loop where i ranges from 1 to n. Initially, i is equal to 1. Inside this loop, we compute the index of a minimal value in the array, from, within the list from i to n. We do this as follows, so we create a variable, minlndex which is initially equal to i. And then we go through all the remaining elements inside this part, I mean through elements from i + 1 to n. And if we find a smaller element we update the variable minlndex. So in the end of this for loop, what we have is that minindex is a position of a minimal element inside the array from i to m. Then we swap this element with the element Ai. Namely, when i is equal to one, what we've done, we've found the minimal element in the well array and we've swapped it with the first element. So now, the first element of our array is in its final position. Then under second iteration of our loop, we do the same actually. We find the minimum value, the position of a minimum value inside the remaining part of our array and put it on the second place. On the sort loop we find the minimum value in this remaining part and put it on the place and so on. So we keep growing the sorted part of our array. So when it would be useful to check the online visualization to see how it goes, so let's do this. This visualization shows how selection sort algorithm performs on a few different datasets. Namely on the random datasets, on a sequence which is nearly sorted. Also on a sequence which is sorted in reversed order. And on a sequence which contains just a few unique elements. So let's run this algorithm and see what happens. So you can see that indeed this algorithm just grows the sorted region, the sorted initial region of our array. So another interesting property is it is revealed by this visualization is the following. So the running time of this algorithm actually does not depend on input data. So it only depends on the size of our initial sequence. The other [INAUDIBLE] time of how algorithm is quadratic and this is not difficult to see right? So what we have is two nested loops. In the outer loop, i ranges from 1 to n. In the inner loop, j ranges from i plus 1 to n, to find a minimum inside the remaining part of our array. So in total we have quadratic number of iterations. At this point however, we should ask ourselves whether our estimate was right in time of the selection, so our algorithm was too pessimistic. And this is whar I mean by this. So recall that we have two nested loops. In the outer loop, i ranges from 1 to n. In the inner loop, g ranges from i + 1 to n. So when i is equal to 1, the number of iterations of the inner loop is n- 1. However, when i is equal to 2, the number of iterations of the inner loop is n- 2, and so on. So when i increases, the number of iterations of the inner loop decreases. So a more accurate estimate for the total number of iterations of the inner loop would be the following, (n- 1) + (n- 2) + (n- 3) and so on. So it is definitely less than n-squared. However we will show this it is equal to n-squared. Namely, this is xx n-squared, and this it is roughly equaled n-square divided by two. The sum is that we need to estimate is called an Arithmetic Series, and there is a known formula for this for this sum. Namely 1 + 2 + 3 +, and so on, n, is equal to n(n+1)/2. And this is how we can prove this formula. Let's just try it, all our n integers in a row, 1, 2, and so on, n. Below them let's write the same set of integers, but in the reverse order. So, n, then n minus 1, and so on, 2, and 1. Then what we get is a row of size 2 by n. Having n columns, and in each column, the sum of the corresponding two integers is equal to n plus 1. Great, so in the first column we have n and one, and in the second column we have two and minus one and so on and in the last column we have n and one. So the sum in each column is equal to n plus one and zero n columns. Which means that the sum of all the numbers in our table is equal to n, when supplied by n plus one. So since this table contains our sum, the sum of the integers from 1 to n twice, we conclude that the sum of all the numbers from 1 to n is equal to n(n+1)/2. Another possibility to find this formula, to see why this formula is correct is to take a rectangle of size n, of dimensions n multiplied by n plus 1. So it's area is equal to n multiplied by n plus one. And to cut it into two parts such as it's shown in the slide, such as the area of each of these two parts is equal to 1 + 2 + and so on n. We're all ready to conclude. So we've just discussed the selection sort algorithm. This algorithm is easy to implement, easy to analyze, and it's running time is n squared, where n is the size of the input sequence. So it sorts the input sequence and array in place. Meaning that it requires almost no extra memory. I mean, all extra memory which is required by this algorithm is only for storing indices, like i, j and m index. There are many other quadratic algorithms, like insertion sort and bubble sort. We're not going to cover them here, and instead, in the next video we will proceed, to do a faster, a faster sort algorithm.
