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 (3.8.2) (judge verdict: Wrong Answer)

from collections import Counter


def num_of_nth_often(lst, nth):
    counted_list = sorted(list(Counter(lst).values()), reverse=True)

    if len(counted_list) < nth:
        return 0
    else:
        return counted_list[nth - 1]


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

v_even = []
v_odd = []

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

if v[0] != v[1]:
    rewrite_num = len(v_even) - num_of_nth_often(v_even, 1) + len(v_odd) - num_of_nth_often(v_odd, 1)
    print(rewrite_num)
else:
    rewrite_num1 = len(v_even) - num_of_nth_often(v_even, 1) + len(v_odd) - num_of_nth_often(v_odd, 2)
    rewrite_num2 = len(v_even) - num_of_nth_often(v_even, 2) + len(v_odd) - num_of_nth_often(v_odd, 1)
    print(max(rewrite_num1, rewrite_num2))