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)

import collections
n = int(input())
nums = list(map(int,input().split()))
a = []
b = []
for i in range(n):
    if i%2 == 0:
        b.append(nums[i])
    else:
        a.append(nums[i])
acol = collections.Counter(a)
bcol = collections.Counter(b)
amost = acol.most_common()[0][1]
bmost = bcol.most_common()[0][1]
if acol.most_common()[0][0] != bcol.most_common()[0][0]:    
    f = amost + bmost
    print(n-f)
elif len(acol.most_common()) != 1 and len(bcol.most_common()) != 1:
    if acol.most_common()[0][1] + bcol.most_common()[1][1] > acol.most_common()[1][1] + bcol.most_common()[0][1]:
        print(n-(acol.most_common()[0][1] + bcol.most_common()[1][1]))
    else:
        print(n-(acol.most_common()[1][1] + bcol.most_common()[0][1]))
elif len(acol.most_common()) == 1 and len(bcol.most_common()) != 1:
    print(n-(acol.most_common()[0][1]+bcol.most_common()[1][1]))
elif len(bcol.most_common()) == 1 and len(acol.most_common()) != 1:
    print(n-(acol.most_common()[1][1]+bcol.most_common()[0][1]))
else:
    print(n//2)