LFEB14A - Editorial

Problem Link:contest, practice

Difficulty: Medium

Pre-requisites: Maths, Inversions, Fenwick Tree

Problem:

We are given two arrays A[] and B[]. At first, we shuffle B[] randomly, then we merge A[] and B[] into array C[] also at random.

Our task is to count the expected number of inversions in C[].

There are N elements in A[] and M elements in B[].

Explanation:

Firstly, as far as the expected value has linear properties, we can represent the answer as AA + BB + AB.

AA is the expected number of inversions between two elements from A[];

BB is the expected number of inversions between two elements from B[];

AB is the expected number of inversions between two elements, such that one of them is from A[] while another is from B[].

Let's focus on each of this summands and calculate them separately.

Calculation of AA:

As far as the elements from A[] preserve their order in C[], AA is a constant and equals to the number of inversions in A[].

It's a well-known problem, you can read about that, for example, here: link

Calculation of BB:

As far as the elements from B[] were shuffled before merging them with A[], they can appear in any order in C[]. By using linear properties of the expected value, we can consider each pair of elements from B[] separetely.

I.e. we consider a pair (X, Y) of integers, assuming that X appears in initial, non-shuffled array B[] earlier than Y. If X equals to Y, then there is no way for this pair to become an inversion. Otherwise(X is different with Y) either X < Y or X > Y. The expected number of inversions of this pair is 1/2.

So, in order to find BB, we should count the number of pairs of elements from B[], such that the numbers from each pair don't equal to each other, and divide it by 2.

Here is a pseudocode, that shows the implementation of the algorithm of calculation BB.

MEM[] is an array, consisting of 100000 zeroes initially
BB = 0
for i from 1 to M do 
begin
    BB = BB + MEM[ B[i] ]
    MEM[ B[i] ] = MEM[ B[i] ] + 1
end
BB = BB / 2

Calculation of AB:

We can use the same logic, as we do for calculating BB: By using linear properties of the expected value, we can consider each pair of elements, such that one of them is from A[] while another is from B[], separetely.

I.e. we consider a pair (X, Y) of integers, assuming that X is a number from A[] while Y is a number from B[].

If X equals to Y, then there is no way for this pair to become an inversion. Otherwise(X is different with Y) either X < Y or X > Y.

In case X < Y the expected number of inversions of this pair equals to i / (N + 1), where i is an index of X in array A[].

In case X > Y the expected number of inversions of this pair equals to (N + 1 - i) / (N + 1), where i is an index of X in array A[].

Why it's so? The point is that the numbers from A[] preserve their initial order in C[]. Then there are (N + 1) different interesting positions for Y in C[]: before A[1], after A[1] and before A[2], after A[2] and before A[3], ..., after A[N]. So, we count only positions that lead pair (X, Y) to an inversion.

How to count AB quickly? We can preprocess numbers from A[], and then for each number from B[] count the expected number of inversions of pairs, where the current B[] is Y.

Here is a pseudocode, that shows the implementation of the algorithm of calculation AB.

MEM1[] is an array, consisting of 100000 zeroes initially
MEM2[] is an array, consisting of 100000 zeroes initially
AB = 0

for i from 1 to N do 
begin
    MEM1[ A[i] ] = MEM1[ A[i] ] + i
    MEM2[ A[i] ] = MEM2[ A[i] ] + (N + 1 - i)
end

for i from 2 to 100000 do
begin
    MEM1[ i ] = MEM1[ i - 1 ] + MEM1[ i ]
end

for i from 99999 to 1 do
begin
    MEM2[ i ] = MEM2[ i + 1 ] + MEM2[ i ]
end

for i from 1 to M do 
begin
    AB = AB + MEM1[ B[i] - 1 ] + MEM2[ B[i] + 1 ]
end

AB = AB / (N + 1)

Some words about the pseudocode.

Here MEM1[i] denotes the sum of indexes of elements from A[], which are not greater than i. Also MEM2[I] denotes the sum of ( N + 1 - index ) of elements from A[], which are not less than i.

Please, look carefully through this section one more time if you don't understand it.

Setter's Solution:link

Tester's Solution:link