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 = [int(i) for i in input().split()]
evens = []
odds = []
for i,num in enumerate(vs):
    if i % 2 == 0:
        evens.append(num)
    elif i % 2 == 1:
        odds.append(num)

def plana():
    if (len(set(vs)) == 2) and (len(set(evens)) == 1) and (len(set(odds)) == 1):
        return 0
    elif (len(set(vs)) == 1):
        return len(evens)
    ec = collections.Counter(evens)
    oc = collections.Counter(odds)

    # check
    if ec.most_common()[0][0] == oc.most_common()[0][0]:
        if ec.most_common()[0][1] >= oc.most_common()[0][1]:
            count = 0
            count += (len(evens) - ec.most_common()[0][1])
            count += (len(odds) - oc.most_common()[1][1])
            return count
        elif ec.most_common()[0][1] < oc.most_common()[0][1]:
            count = 0
            count += (len(evens) - ec.most_common()[1][1])
            count += (len(odds) - oc.most_common()[0][1])
            return count

    count = 0
    count += (len(evens) - ec.most_common()[0][1])
    count += (len(odds) - oc.most_common()[0][1])
    return count

print(int(plana()))