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)

#!/mnt/c/Users/moiki/bash/env/bin/python
n = int(input())
v = list(map(int, input().split()))
vodd = []
dodd = {}
veve = []
deve = {}

kind = 0
inst_kinds = {}
for e,i in enumerate(v):
    if e % 2 == 0:
        # veve.append(i)
        if not i in deve:
            deve[i] = 1
        else:
            deve[i] += 1
    else:
        # vodd.append(i)
        if not i in dodd:
            dodd[i] = 1
        else:
            dodd[i] += 1

for k in deve:
    veve.append((deve[k], k))
veve.sort()
veve.reverse()
for k in dodd:
    vodd.append((dodd[k], k))
vodd.sort()
vodd.reverse()
from functools import reduce
from operator import add

# sum = reduce(add, [k[0] for k in vodd])
sum = n
# sum += reduce(add, [k[0] for k in veve])

# print("beforesum", sum)
# print(vodd, veve)
if vodd[0][0] >= veve[0][0]:
    sum -= vodd[0][0]
    if veve[0][1] == vodd[0][1] and len(veve) >= 2:
        sum -= veve[1][0]
    elif not veve[0][1] == vodd[0][1]:
        sum -= veve[0][0]

elif vodd[0][0] < veve[0][0]:
    sum -= veve[0][0]
    if veve[0][1] == vodd[0][1] and len(vodd) >= 2:
        sum -= vodd[1][0]
    elif not  veve[0][1] == vodd[0][1]:
        sum -= vodd[0][0]

print(sum)