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)

#encoding utf-8

n = int(input())
v = list(int(i) for i in input().split())

even_v = []
odd_v = []

for i in range(n):
    if i%2 == 0:
        odd_v.append(v[i])
    else:
        even_v.append(v[i])

odd_dic = {}
even_dic = {}
for i in range(len(odd_v)):
    if odd_v[i] in odd_dic.keys():
        odd_dic[odd_v[i]] += 1
    else:
        odd_dic[odd_v[i]] = 1

for i in range(len(even_v)):
    if even_v[i] in even_dic.keys():
        even_dic[even_v[i]] += 1
    else:
        even_dic[even_v[i]] = 1


for k, v in sorted(even_dic.items(), key=lambda x: -x[1]):pass
for k, v in sorted(odd_dic.items(), key=lambda x: -x[1]):pass

# odd_dic = sorted(odd_dic.items(),key=lambda x:x[1])
# even_dic = sorted(even_dic.items(),key=lambda x:x[1])

odd_keys = list(odd_dic.keys())
even_keys = list(even_dic.keys())

if odd_keys[0] != even_keys[0]:
    answer = n-int(odd_dic[odd_keys[0]])-int(even_dic[even_keys[0]])
else:
    if len(odd_keys) == 1 and len(even_keys) == 1:
        answer = len(odd_v)
    else:
        ans1 = n-int(odd_dic[odd_keys[0]])-int(even_dic[even_keys[1]])
        ans2 = n-int(odd_dic[odd_keys[1]])-int(even_dic[even_keys[0]])
        answer = min(ans1, ans2)
print(int(answer))