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: Time Limit Exceeded)

import heapq

n = int(input())
v = list(map(int,input().split()))
a_odd = []
a_even = []
check_a_odd = []
check_a_even = []
count_a_odd = []
count_a_even = []

for i in range(len(v)):
    if i%2 == 0:
        a_odd.append(v[i])
    else:
        a_even.append(v[i])

#print(a_odd)
#print(a_even)

for j in range(len(a_odd)):
    if a_odd[j] not in check_a_odd:
        heapq.heappush(count_a_odd,[-1*a_odd.count(a_odd[j]),a_odd[j]])
        check_a_odd.append(a_odd[j])
    if a_even[j] not in check_a_even:
        heapq.heappush(count_a_even,[-1*a_even.count(a_even[j]),a_even[j]])
        check_a_even.append(a_even[j])

count_a_odd.append([0,0])
count_a_even.append([0,0])

#print(count_a_odd)
#print(count_a_even)

if int(count_a_odd[0][1]) != int(count_a_even[0][1]):
    print(n + int(count_a_odd[0][0]) + int(count_a_even[0][0]))
else:
    ans1 = n + int(count_a_odd[0][0]) + int(count_a_even[1][0])
    ans2 = n + int(count_a_odd[1][0]) + int(count_a_even[0][0])
    print(min(ans1,ans2))