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: Wrong Answer)

# -*- coding: utf-8 -*-
from collections import Counter

n = int(input().strip())
v_list = list(map(int, input().rstrip().split()))
#-----

v_odd = [ v_list[i] for i in range(0,len(v_list),2) ]
v_even  = [ v_list[i] for i in range(1,len(v_list),2) ]

sort_cnt_v_odd  = sorted( Counter(v_odd).items(), key=lambda x:x[1], reverse=True)
sort_cnt_v_even = sorted(Counter(v_even).items(), key=lambda x:x[1], reverse=True)


ans=0
# カウント数最大の数値が、v_oddとv_evenで異なるとき
# v_oddとv_evenそれぞれを、1種類の数値に書き換え
if sort_cnt_v_odd[0][0] != sort_cnt_v_even[0][0]:
    ans +=  len(v_odd) -  sort_cnt_v_odd[0][1]
    ans += len(v_even) - sort_cnt_v_even[0][1]
else:
    # v_oddとv_evenそれぞれで、数値が1種類のとき
    # v_oddを全書き換え
    if len(sort_cnt_v_odd)==len(sort_cnt_v_even)==1:
        ans = sort_cnt_v_odd[0][1]
        
    # v_oddの、数値が1種類のとき、v_evenは2番目に多いカウント数の数値に書き換え
    elif len(sort_cnt_v_odd)==1 and len(sort_cnt_v_even)!=1:
        ans = sort_cnt_v_odd[0][1] + sort_cnt_v_even[1][1]
        
    # v_evenの、数値が1種類のとき
    elif len(sort_cnt_v_odd)!=1 and len(sort_cnt_v_even)==1:
        ans = sort_cnt_v_odd[1][1] + sort_cnt_v_even[0][1]
        
    # 以下2パターンの少ない方を採用
    #   v_odd　 最大カウント数、v_even　2番目のカウント数
    #   v_even　最大カウント数、v_odd　 2番目のカウント数
    elif len(sort_cnt_v_odd)!=1 and len(sort_cnt_v_even)!=1:
        tmp1 = sort_cnt_v_odd[0][1] + sort_cnt_v_even[1][1]
        tmp2 = sort_cnt_v_odd[1][1] + sort_cnt_v_even[0][1]
        ans = min(tmp1,tmp2)

print(ans)