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)

import collections
N=int(input())
A=list(map(int,input().split()))

A1=[]
A2=[]
for i in range(len(A)):
    if i%2==0:
        A1.append(A[i])
    else:
        A2.append(A[i])
        
B1=collections.Counter(A1)
B2=collections.Counter(A2)


if B1.most_common(2)[0][1]==B2.most_common(2)[0][1] and B1.most_common(2)[0][1]==int(N/2):
    print(int(N/2))
else:
    
    if B1.most_common(1)==B2.most_common(1):
        if B1.most_common(2)[1][1]>=B2.most_common(2)[1][1]:
            ans=N-B1.most_common(2)[1][1]-B2.most_common(2)[0][1]
        else:
            ans=N-B2.most_common(2)[1][1]-B1.most_common(2)[0][1]
    elif B1.most_common(2)[0][0]==B2.most_common(2)[0][0]:
        if B1.most_common(2)[0][1]>=B1.most_common(2)[0][1]:
            ans=N-B1.most_common(2)[0][1]-B2.most_common(2)[1][1]
        else:
            ans=N-B2.most_common(2)[0][1]-B1.most_common(2)[1][1]
    else:
        ans=N-B1.most_common(2)[0][1]-B2.most_common(2)[0][1]
    print(ans)