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)

import collections


n = int(input())
v = list(map(int, input().split()))
odd = []
even = []
for i in range(int(n/2)):
    even.append(v[2*i])
    odd.append(v[2*i + 1])
evencount = collections.Counter(even)
oddcount = collections.Counter(odd)
 
if evencount.most_common()[0][0] != oddcount.most_common()[0][0]:
    ans = n - evencount.most_common()[0][1] - oddcount.most_common()[0][1]
else:
    if evencount.most_common()[0][1] > oddcount.most_common()[0][1]:
        if len(oddcount)!=1:
            ans = n - evencount.most_common()[0][1] - oddcount.most_common()[1][1]
        else:
            ans = n - evencount.most_common()[0][1]
    else:
        if len(evencount)!=1:
            ans = n - evencount.most_common()[1][1] - oddcount.most_common()[0][1]
        else:
            ans = n - oddcount.most_common()[0][1]

    
print(ans)