Problem:
Score : 300 points

Problem Statement

A sequence a_1,a_2,... ,a_n is said to be /\/\/\/ when the following conditions are satisfied:

For each i = 1,2,..., n-2, a_i = a_{i+2}.

Exactly two different numbers appear in the sequence.

You are given a sequence v_1,v_2,...,v_n whose length is even.
We would like to make this sequence /\/\/\/ by replacing some of its elements.
Find the minimum number of elements that needs to be replaced.

Constraints

2 \leq n \leq 10^5

n is even.

1 \leq v_i \leq 10^5

v_i is an integer.

Input

Input is given from Standard Input in the following format:

n
v_1 v_2 ... v_n

Output

Print the minimum number of elements that needs to be replaced.

Sample Input 1

4
3 1 3 2

Sample Output 1

1

The sequence 3,1,3,2 is not /\/\/\/, but we can make it /\/\/\/ by replacing one of its elements: for example, replace the fourth element to make it 3,1,3,1.

Sample Input 2

6
105 119 105 119 105 119

Sample Output 2

0

The sequence 105,119,105,119,105,119 is /\/\/\/.

Sample Input 3

4
1 1 1 1

Sample Output 3

2

The elements of the sequence 1,1,1,1 are all the same, so it is not /\/\/\/.

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

let n = Scanf.scanf " %d" (+) 0
let vs = Array.init n @@ fun _ -> Scanf.scanf " %d" (+) 0
let css = Array.make_matrix 2 100001 0
let ms = ref [|0, 0; 0, 0|]
let e0, e1, o0, o1 = Array.iteri (fun i v -> let c = css.(i mod 2).(v) + 1 in css.(i mod 2).(v) <- c; !ms.(i mod 2) <- max !ms.(i mod 2) (c, v)) vs;
  List.iter (fun i -> Array.sort (fun x y -> compare y x) css.(i)) [0; 1];
  css.(0).(0), css.(0).(1), css.(1).(0), css.(1).(1)
let _ = Printf.printf "%d\n" @@ if snd !ms.(0) = snd !ms.(1) then n - max (e0 + o1) (e1 + o0) else n - e0 - o0