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 Haskell (GHC 7.10.3) (judge verdict: Wrong Answer)

import Data.List

main :: IO ()
main = do
  _ <- getLine
  vs <- map read . words <$> getLine
  print $ solve vs

solve :: [Int] -> Int
solve vs = solve' (count as) (count bs)
  where
    (as, bs) = deal vs

deal :: [Int] -> ([Int], [Int])
deal [] = ([], [])
deal (x:y:zs) = (x:xs, y:ys)
  where
    (xs, ys) = deal zs

solve' :: [(Int, Int)] -> [(Int, Int)] -> Int
solve' [a0] [b0]
  | ka0 == kb0 = va0 `min` vb0
  | ka0 /= kb0 = 0
  where
    (ka0, va0) = a0
    (kb0, vb0) = b0
solve' [a0] (b0:b1:bs)
  | ka0 == kb0 = vb0 + bsum
  | ka0 /= kb0 = vb1 + bsum
  where
    (ka0, va0) = a0
    (kb0, vb0) = b0
    (kb1, vb1) = b1
    bsum = sum (map snd bs)
solve' (a0:a1:as) [b0]
  | ka0 == kb0 = va0 + asum
  | ka0 /= kb0 = va1 + asum
  where
    (ka0, va0) = a0
    (kb0, vb0) = b0
    (ka1, va1) = a1
    asum = sum (map snd as)
solve' (a0:a1:as) (b0:b1:bs)
  | ka0 /= kb0             = va1 + asum + vb1 + bsum
  | va0 + vb1 >= va1 + vb0 = va1 + asum + vb0 + bsum
  | va0 + vb1 <  va1 + vb0 = va1 + asum + vb1 + bsum
  where
    (ka0, va0) = a0
    (kb0, vb0) = b0
    (ka1, va1) = a1
    (kb1, vb1) = b1
    asum = sum (map snd as)
    bsum = sum (map snd bs)

count :: [Int] -> [(Int, Int)]
count = reverse . sortOn snd . map (\xs -> (head xs, length xs)) . group . sort