A stable bubble week

TopCoder Open 2020 Round 2A was the main event of Jul 6 - Jul 12 week (problems, results, top 5 on the left, parallel round results, analysis). Four contestants solved all three problems in the main round, which is even more impressive given that nobody managed to do that in the parallel round, despite that fact that the strongest contestants who qualified directly to Round 4 were competing there. Congratulations to all four, and especially to Kriii on the win!

In my previous summary, I have mentioned a Codeforces problem: you are given an array a with at most 1000 elements. Then we write down all pairs of positions that form an inversion: pairs (u,v) such that u<v and a_u>a_v, getting a long list of all those pairs. Now we want to treat this list as a sorting program: for every pair in the list, we will swap the elements on the corresponding positions. Our goal is to make this program actually sort our array. We are allowed to put the elements of the list in arbitrary order (but we must have all pairs that form an inversion in the starting array exactly once).

If we were allowed to swap any pair that forms an inversion in the current state of the array, then the bubble sort algorithm would work, as it only swaps adjacent elements that form an inversion. However, we can only use the inversions from the initial array (and must use them all).

In order to achieve this, we need to find an algorithm that tries to keep most existing inversions unchanged. Let's do the following: first, we find the maximum number. Then, we find the second highest number, and sort them (within their places). Then, we add the third highest number and sort the three numbers within their places, using up all their inversions, and so on. This way, whenever we process a new number, the only thing that happened to the array is that the higher numbers got reordered, so the inversions involving the new number stay unchanged!

The only remaining step is to learn how to put the new number into its correct place using all its inversions. This is equivalent to putting 1 into the correct place given the array 2 3 4 ... k 1 (k+1) ... n. The following sequence of swaps does the job: swap 2 and 1, then swap 3 and 2 (which is in the original position of 1 now), then swap 4 and 3, and so on until swapping k and k-1.

Thanks for reading, and check back for more!