Problem:
Score : 400 points

Problem Statement

You are given N integers; the i-th of them is A_i.
Find the maximum possible sum of the absolute differences between the adjacent elements after arranging these integers in a row in any order you like.

Constraints

2 \leq N \leq 10^5

1 \leq A_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1
:
A_N

Output

Print the maximum possible sum of the absolute differences between the adjacent elements after arranging the given integers in a row in any order you like.

Sample Input 1

5
6
8
1
2
3

Sample Output 1

21

When the integers are arranged as 3,8,1,6,2, the sum of the absolute differences between the adjacent elements is |3 - 8| + |8 - 1| + |1 - 6| + |6 - 2| = 21. This is the maximum possible sum.

Sample Input 2

6
3
1
4
1
5
9

Sample Output 2

25

Sample Input 3

3
5
5
1

Sample Output 3

8

Solution in OCaml (4.02.3) (judge verdict: Accepted)

open Printf
open Scanf

let id x = x
let pair x y = x, y

let n = scanf "%d" id
let vs = Array.init n (fun _ -> scanf " %d" id)

let () = Array.sort (fun x y -> x - y) vs

let odd1 vs =
  let mid = n / 2 in
  let up = ref (n - 1) in
  let lo = ref 0 in
  let r1 = ref 0 in
  let p = ref vs.(mid) in
  for i = 0 to n / 2 - 1 do
    r1 := !r1 + abs (!p - vs.(!up));
    p := vs.(!up);
    decr up;
    r1 := !r1 + abs (!p - vs.(!lo));
    p := vs.(!lo);
    incr lo
  done;
  !r1
  
let odd2 vs =
  let mid = n / 2 in
  let up = ref (n - 1) in
  let lo = ref 0 in
  let r1 = ref 0 in
  let p = ref vs.(mid) in
  for i = 0 to n / 2 - 1 do
    r1 := !r1 + abs (!p - vs.(!lo));
    p := vs.(!lo);
    incr lo;
    r1 := !r1 + abs (!p - vs.(!up));
    p := vs.(!up);
    decr up
  done;
  !r1

let even1 vs =
  let mid = n / 2 in
  let up = ref (n - 1) in
  let lo = ref 0 in
  let r1 = ref 0 in
  let p = ref vs.(mid) in
  for i = 0 to n / 2 - 2 do
    r1 := !r1 + abs (!p - vs.(!lo));
    p := vs.(!lo);
    incr lo;
    r1 := !r1 + abs (!p - vs.(!up));
    p := vs.(!up);
    decr up
  done;
  r1 := !r1 + abs (!p - vs.(mid - 1));
  !r1

let even2 vs =
  let mid = n / 2 - 1 in
  let up = ref (n - 1) in
  let lo = ref 0 in
  let r1 = ref 0 in
  let p = ref vs.(mid) in
  for i = 0 to n / 2 - 2 do
    r1 := !r1 + abs (!p - vs.(!up));
    p := vs.(!up);
    decr up;
    r1 := !r1 + abs (!p - vs.(!lo));
    p := vs.(!lo);
    incr lo;
  done;
  r1 := !r1 + abs (!p - vs.(mid + 1));
  !r1

let () =
  if n mod 2 = 0 then begin
    let e1 = even1 vs in
    let e2 = even2 vs in
    printf "%d\n" (max e1 e2)
  end else begin
    let o1 = odd1 vs in
    let o2 = odd2 vs in
    printf "%d\n" (max o1 o2)
  end