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())
vs = list(map(int, input().split()))
odd = vs[::2]
even = vs[1::2]
c_odd = collections.Counter(odd)
c_even = collections.Counter(even)
v_odd = c_odd.most_common(1)[0][0]
v_even = c_even.most_common(1)[0][0]
if v_odd == v_even:
    if len(c_odd) == 1 and len(c_even) == 1:
        same = n//2
    elif len(c_odd) == 1 and len(c_even) != 1:
        same = n//2 + c_even.most_common(2)[1][1]
    elif len(c_odd) != 1 and len(c_even) == 1:        
        same = n//2 + c_odd.most_common(2)[1][1]
    else:
        v2_odd = c_odd.most_common(2)
        v2_even = c_even.most_common(2)    
        if v2_odd[1][1] > v2_even[1][1]:
            same = c_odd.most_common(1)[0][1] + v2_even[1][1]
        else:
            same =  v2_odd[1][1]+ c_even.most_common(1)[0][1]
else:
    same =  c_odd.most_common(1)[0][1] + c_even.most_common(1)[0][1]
print(n - same)