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)

#gcdは3.4.3だとfractionsにある
from fractions import gcd

from math import factorial as f

from math import ceil,floor,sqrt

from sys import exit

from copy import deepcopy
import numpy as np

import bisect
import re


ii = lambda : int(input())
mi = lambda : map(int,input().split())
li = lambda : list(map(int,input().split()))



def main():
    n = int(input())
    v = list(map(int,input().split()))

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

    ocount = [0 for i in range(100005)]
    ecount = [0 for i in range(100005)]

    for i in odd:
        ocount[i] += 1
    for i in even:
        ecount[i] += 1

    om = np.argmax(ocount)
    em = np.argmax(ecount)

    if len(set(v)) == 1:
        print(n//2)
        exit()

    if not om == em:
        print(n-(ocount[om]+ecount[em]))
        exit()

    if ocount[om] > ecount[em]:
        ecount[em] = 0
        em = np.argmax(ecount)
        print(n-(ocount[om]+ecount[em]))
    elif ecount[em] > ocount[om]:
        ocount[em] = 0
        om = np.argmax(ocount)
        print(n-(ocount[om]+ecount[em]))
    else:
        oc = ocount[om]
        ecount[em] = 0
        em = np.max(ecount)
        ocount[em] = 0
        om = np.max(ocount)

        print(n-(oc + max(om,em)))