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: Runtime Error)

#111c
import collections
import copy

n = int(input()) #2≤n≤10**5 偶数
v_list = [int(e) for e in input().split()]

#数列 a1,a2,...,anが以下の条件を満たすとき、 /\/\/\/ と呼ぶことにします。

#・各 i=1,2,...,n−2について、ai=ai+2 
#・数列に現れる数はちょうど 2種類

#偶数長の数列 v1,v2,...,vnが与えられます。 
#要素をいくつか書き換えることでこの数列を /\/\/\/ にしたいです。 書き換える要素の数は最小でいくつになるか求めてください。

#一番多く出てくる文字を求め、書き換えない数字を決める
v_list_odd = [v_list[i] for i in range(n) if i%2==1]
v_list_even = [v_list[i] for i in range(n) if i%2==0]

#print(v_list_odd)
#print(v_list_even)

c = collections.Counter(v_list_odd)
v_list_odd_counter = copy.copy(c)
if len(c)>1:
    fix_picklist_for_v_list_odd = [((c.most_common(2))[0])[0],((c.most_common(2))[0])[1],((c.most_common(2))[1])[0],((c.most_common(2))[1])[1]]
else:
    fix_char_for_v_list_odd = fix_picklist_for_v_list_odd = [((c.most_common(2))[0])[0],((c.most_common(2))[0])[1]]
    
c = collections.Counter(v_list_even)
v_list_even_counter = copy.copy(c)
if len(c)>1:
    fix_picklist_for_v_list_even = [((c.most_common(2))[0])[0],((c.most_common(2))[0])[1],((c.most_common(2))[1])[0],((c.most_common(2))[1])[1]]
else:
    fix_picklist_for_v_list_even = [((c.most_common(2))[0])[0],((c.most_common(2))[0])[1]]
    
#固定文字の決定
#1文字目が重複していなければ、両方とも出現回数最多を固定する
if fix_picklist_for_v_list_even[0] != fix_picklist_for_v_list_odd[0]:
    fix_char_for_v_list_even = fix_picklist_for_v_list_even[0]
    fix_char_for_v_list_odd = fix_picklist_for_v_list_odd[0]

#1文字目が重複している場合
#両方1文字しかなければ片方のみ
elif len(fix_picklist_for_v_list_even) == 2 and len(fix_picklist_for_v_list_odd) == 2:
    fix_char_for_v_list_even = fix_picklist_for_v_list_even[0]
    fix_char_for_v_list_odd = "!"
#片方が1文字ならば、そうでない方は2文字目を使う。
elif len(fix_picklist_for_v_list_even) == 2 and len(fix_picklist_for_v_list_odd) != 2:
    fix_char_for_v_list_even = fix_picklist_for_v_list_even[0]
    fix_char_for_v_list_odd = fix_picklist_for_v_list_odd[2]
elif len(fix_picklist_for_v_list_even) != 2 and len(fix_picklist_for_v_list_odd) == 2:
    fix_char_for_v_list_even = fix_picklist_for_v_list_even[2]
    fix_char_for_v_list_odd = fix_picklist_for_v_list_odd[0]
#両方2文字あるなら、固定パターンの合計が多い方を採用
elif fix_picklist_for_v_list_even[1]+fix_picklist_for_v_list_odd[3] > fix_picklist_for_v_list_even[3]+fix_picklist_for_v_list_odd[1]:
    fix_char_for_v_list_even = fix_picklist_for_v_list_even[0]
    fix_char_for_v_list_odd = fix_picklist_for_v_list_odd[2]
elif  fix_picklist_for_v_list_even[1]+fix_picklist_for_v_list_odd[3]= fix_picklist_for_v_list_even[3]+fix_picklist_for_v_list_odd[1]:
    fix_char_for_v_list_even = fix_picklist_for_v_list_even[2]
    fix_char_for_v_list_odd = fix_picklist_for_v_list_odd[0]

#print(fix_char_for_v_list_odd)
#print(fix_char_for_v_list_even)
        
#書き換え要素数をカウント
ans_odd = n/2 - v_list_odd_counter[fix_char_for_v_list_odd]
ans_even = n/2 - v_list_even_counter[fix_char_for_v_list_even]
print(int(ans_odd + ans_even))