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)

if __name__ == '__main__':
    n = int(input())
    vs = list(map(int, input().split()))

    even = vs[::2]
    odd = vs[1::2]

    from collections import Counter

    result = 0
    #evenの処理
    ce = Counter(even)
    emode = ce.most_common(1)[0]

    oe = Counter(odd)
    omode = oe.most_common(1)[0]
    if emode[0] == omode[0]:
        if emode[1] > omode[1]:
            for i in range(len(even)):
                if even[i] != emode[0]:
                    result += 1
            if len(oe.most_common()) == 1:
                omode = -1
                result += len(odd)
            else:
                omode = oe.most_common(2)[1][0]
                for i in range(len(odd)):
                    if odd[i] != omode:
                        result += 1

        else:
            for i in range(len(odd)):
                if odd[i] != omode[0]:
                    result += 1
            if len(ce.most_common()) == 1:
                emode = -1
                result += len(even)
            else:
                emode = ce.most_common(2)[1][0]
                for i in range(len(even)):
                    if even[i] != emode:
                        result += 1
    else:
        for i in range(len(even)):
            if even[i] != emode[0]:
                result += 1
        for i in range(len(odd)):
            if odd[i] != omode[0]:
                result += 1

    print(result)