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 PyPy3 (7.3.0) (judge verdict: Runtime Error)

from collections import Counter
n = int(input())
arr = list(map(int, input().split()))

odd_list =[]
num_list =[]

for i in range(n):
    if i%2 == 0:
        odd_list.append(arr[i])
    if i%2 != 0:
        num_list.append(arr[i])

odd_count = Counter(odd_list)
num_count = Counter(num_list)
odd1 = odd_count[odd_count.most_common()[0][0]]
num1 = num_count[num_count.most_common()[0][0]]



if len(set(arr)) == 1:
    print(int(n/2))
    exit()

if len(set(num_list))==1 and len(set(odd_list))==1:
    print(0)
    exit()

if len(set(odd_count)) == 1 and len(set(num_count)) != 1:
    num2 = num_count[num_count.most_common()[1][0]]
    if odd_list[0] == num_count.most_common()[0][0]:
        print(int(n/2)-num2)
        exit()
    if odd_list[0] != num_count.most_common()[0][0]:
        print(int(n/2)-num1)
        exit()

if len(set(odd_count)) != 1 and len(set(num_count)) == 1:
    odd2 = odd_list[odd_count.most_common()[1][0]]
    if num_list[0] == odd_count.most_common()[0][0]:
        print(int(n/2)-odd2)
        exit()
    if num_list[0] != odd_count.most_common()[0][0]:
        print(int(n/2)-odd1)
        exit()

odd2 = odd_count[odd_count.most_common()[1][0]]
num2 = num_count[num_count.most_common()[1][0]]

if odd_count.most_common()[0][0] != num_count.most_common()[0][0]:
    print(n-odd1-num1)

if odd_count.most_common()[0][0] == num_count.most_common()[0][0]:
    print(min(n-odd1-num2, n-odd2-num1))