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)

n=int(input())
vlis=[int(i) for i in input().split()]

# 始点から5(6-1)番目の要素を2個ごとに取り出す
# a[:6:2] #[0, 2, 4]
# 7番目から終点までの要素を2個ごとに取り出す
# a[7::2] #[7, 9]
odd=vlis[0::2]
even=vlis[1::2]



import collections
odd_max=collections.Counter(odd).most_common(2)
even_max=collections.Counter(even).most_common(2)

# 一番多い数の数字が偶奇で同じ場合
if odd_max[0][0]==even_max[0][0]:
    # 全てが同じ数字の場合
    if len(odd_max)==1 and len(even_max)==1:
        print(int(len(vlis)/2))
    # 偶数列が全部同じ値だった時は奇数列は２番目多い数の数字に合わせる
    elif len(even_max)==1:
        print(len(vlis)-even_max[0][1]-odd_max[1][1])
    # 奇数列が全部同じ値だった時は偶数列は２番目多い数の数字に合わせる
    elif len(odd_max)==1:
        print(len(vlis)-odd_max[0][1]-even_max[1][1])
    else:

        if odd_max[0][1]+even_max[1][1]>=odd_max[1][1]+even_max[0][1]:
            print(len(vlis)-odd_max[0][1]-even_max[1][1])
        else:
            print(len(vlis)-odd_max[1][1]-even_max[0][1])
else:
    print(len(vlis)-odd_max[0][1]-even_max[0][1])