How to find number of expected swaps in bubble sort in better than O(n^2) time

Question

I am stuck on problem http://www.codechef.com/JULY12/problems/LEBOBBLE Here it is required to find number of expected swaps.

I tried an O(n^2) solution but it is ti

Answer 1

Use divide and conquer

divide: size of sequence n to two lists of size n/2 conquer: count recursively two lists combine: this is a trick part (to do it in linear time)

combine use merge-and-count. Suppose the two lists are A, B. They are already sorted. Produce an output list L from A, B while also counting the number of inversions, (a,b) where a is-in A, b is-in B and a > b.

The idea is similar to "merge" in merge-sort. Merge two sorted lists into one output list, but we also count the inversion.

Everytime a_i is appended to the output, no new inversions are encountered, since a_i is smaller than everything left in list B. If b_j is appended to the output, then it is smaller than all the remaining items in A, we increase the number of count of inversions by the number of elements remaining in A.

merge-and-count(A,B) ; A,B two input lists (sorted) ; C output list ; i,j current pointers to each list, start at beginning ; a_i, b_j elements pointed by i, j ; count number of inversion, initially 0

while A,B != empty append min(a_i,b_j) to C if b_j < a_i count += number of element remaining in A j++ else i++ ; now one list is empty append the remainder of the list to C return count, C

With merge-and-count, we can design the count inversion algorithm as follows:

sort-and-count(L) if L has one element return 0 else divide L into A, B (rA, A) = sort-and-count(A) (rB, B) = sort-and-count(B) (r, L) = merge-and-count(A,B) return r = rA+rB+r, L

T(n) = O(n lg n)