Question

This problem is taken from interviewstreet.com

Given array of integers Y=y1,…,yn, we have n line segments such that
endpoints of segment i are (i, 0) and (i, yi). Imagine that from the
top of each segment a horizontal ray is shot to the left, and this ray
stops when it touches another segment or it hits the y-axis. We
construct an array of n integers, v1, …, vn, where vi is equal to
length of ray shot from the top of segment i. We define V(y1, …, yn)
= v1 + … + vn.

For example, if we have Y=[3,2,5,3,3,4,1,2], then v1, …, v8 =
[1,1,3,1,1,3,1,2], as shown in the picture below:

For each permutation p of [1,…,n], we can calculate V(yp1, …,
ypn). If we choose a uniformly random permutation p of [1,…,n], what
is the expected value of V(yp1, …, ypn)?

Input Format

First line of input contains a single integer T (1 <= T <= 100). T
test cases follow.

First line of each test-case is a single integer N (1 <= N <= 50).
Next line contains positive integer numbers y1, …, yN separated by a
single space (0 < yi <= 1000).

Output Format

For each test-case output expected value of V(yp1, …, ypn), rounded
to two digits after the decimal point.

Sample Input

6
3
1 2 3
3
3 3 3
3
2 2 3
4
10 2 4 4
5
10 10 10 5 10
6
1 2 3 4 5 6

Sample Output

4.33
3.00
4.00
6.00
5.80
11.15

Explanation

Case 1: We have V(1,2,3) = 1+2+3 = 6, V(1,3,2) = 1+2+1 = 4, V(2,1,3) =
1+1+3 = 5, V(2,3,1) = 1+2+1 = 4, V(3,1,2) = 1+1+2 = 4, V(3,2,1) =
1+1+1 = 3. Average of these values is 4.33.

Case 2: No matter what the permutation is, V(yp1, yp2, yp3) = 1+1+1 =
3, so the answer is 3.00.

Case 3: V(y1 ,y2 ,y3)=V(y2 ,y1 ,y3) = 5, V(y1, y3, y2)=V(y2, y3, y1) =
4, V(y3, y1, y2)=V(y3, y2, y1) = 3, and average of these values is
4.00.

A naive solution to the problem will run forever for N=50. I believe that the problem can be solved by independently calculating a value for each stick. I still need to know if there is any other efficient approach for this problem. On what basis do we have to independently calculate value for each stick?

Answer 1

As you correctly, noted we can solve problem independently for each stick.

Let F(i, len) is number of permutations, that ray from stick i is exactly len.
Then answer is

(Sum(by i, len) F(i,len)*len)/(n!)

All is left is to count F(i, len). Let a(i) be number of sticks j, that y_j<=y_i. b(i) – number of sticks, that b_j>b_i.

In order to get ray of length len, we need to have situation like this.

B, l...l, O  
   len-1 times

Where O – is stick #i. B – is stick with bigger length, or beginning. l – is stick with heigth, lesser then ith.

This gives us 2 cases:
1) B is the beginning, this can be achieved in P(a(i), len-1) * (b(i)+a(i)-(len-1))! ways.
2) B is bigger stick, this can be achieved in P(a(i), len-1)*b(i)*(b(i)+a(i)-len)!*(n-len) ways.

edit: corrected b(i) as 2nd term in (mul)in place of a(i) in case 2.