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: Accepted)

from collections import Counter
n = int(input())
A = list(map(int,input().split()))
E = [A[i] for i in range(n) if i%2 == 1]
O = [A[i] for i in range(n) if i%2 == 0]
e = Counter(E).most_common(2)
o = Counter(O).most_common(2)
if e[0][0] != o[0][0]: ##Eの最頻値とOの最頻値が違う
    ans = (n//2-e[0][1])+(n//2-o[0][1])
    print(ans)
    ##以下、EとOの最頻値が同じ時の処理を書く
elif len(e)>=2 and len(o)>=2:
    ans = min((n//2-e[0][1])+(n//2-o[1][1]), (n//2-e[1][1])+(n//2-o[0][1]))
    print(ans)
else:
    print(n//2)