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 Python (3.4.3) (judge verdict: Accepted)

import numpy as np
from collections import Counter
n = int(input())
As = input().split()
As = [int(A) for A in As]

A_odd = As[::2]         #奇数列
A_even = As[1:][::2]    #偶数列

# 最頻の要素を並べる
odd_pair = Counter(A_odd).most_common(2)
even_pair = Counter(A_even).most_common(2)

if odd_pair[0][0] != even_pair[0][0]:               # TOPの数字が違う
    N = n-odd_pair[0][1]-even_pair[0][1]
else:                                               # TOPの数字が一緒
    if len(odd_pair) == 1 and len(even_pair) == 1:  # 文字列がそもそも1種類
        N = n/2
    elif len(odd_pair) == 1:                        # oddが1種類
        N = n-even_pair[1][1]
    elif len(even_pair) == 1:                       # evenが1種類
        N = n-odd_pair[1][1] 
    elif odd_pair[1][1] > even_pair[1][1]:          # oddの2番目のが多い
        N = n-even_pair[0][1]-odd_pair[1][1]
    else:
        N = n-odd_pair[0][1]-even_pair[1][1]        # oddの1番目のが多い
print(int(N))