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: Runtime Error)

nums = int(input())
odds = []
even = []
items = [int(i) for i in input().split(' ')]
for i,data in enumerate(items):
  if i%2 == 0:
    even.append(data)
  else:
    odds.append(data)

def get_frequent_data(dataset):
	return max(set(dataset), key=dataset.count)

def get_second_frequent(dataset):
  first = get_frequent_data(dataset)
  delete_first = [ i for i in dataset if i != first]
  second = get_frequent_data(delete_first)
  return second

most_frequent_odds = get_frequent_data(odds)
most_frequent_even = get_frequent_data(even)
second_frequent_odds = get_second_frequent(odds)
second_frequent_even = get_second_frequent(even)

if most_frequent_odds != most_frequent_even:
  print(len([i for i in odds if i != most_frequent_odds])+
  		len([i for i in even if i != most_frequent_even]))
else:
  even_len = len([i for i in odds if i == most_frequent_even])
  odds_len = len([i for i in odds if i == most_frequent_odds])
  if even_len > odds_len:
  	print(len([i for i in odds if i != second_frequent_odds])+
  		len([i for i in even if i != most_frequent_even]))
  elif even_len < odds_len:
  	print(len([i for i in odds if i != most_frequent_odds])+
  		len([i for i in even if i != second_frequent_even]))
  else:
    print(len([i for i in odds if i != second_frequent_odds])+
  		len([i for i in even if i != most_frequent_even]))