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)

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

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

freq_e = {}
freq_o = {}
for j in range(int(n / 2)):
  if (str(e[j]) in freq_e):
    freq_e[str(e[j])] += 1
  elif not(str(e[j]) in freq_e):
    freq_e[str(e[j])] = 1
  if (str(o[j]) in freq_o):
    freq_o[str(o[j])] += 1
  elif not(str(o[j]) in freq_o):
    freq_o[str(o[j])] = 1

freq_e_sorted = sorted(freq_e.items(), key = lambda x: -x[1])
freq_o_sorted = sorted(freq_o.items(), key = lambda x: -x[1])
if freq_e_sorted[0][0] != freq_o_sorted[0][0]:
  ans = n - freq_e_sorted[0][1] - freq_o_sorted[0][1]
elif len(freq_e_sorted) == 1 and len(freq_o_sorted) == 1:
  ans = freq_e_sorted[0][1]
elif len(freq_e_sorted) == 1 or freq_e_sorted[0][1] + freq_o_sorted[1][1] > freq_e_sorted[1][1] + freq_o_sorted[0][1]:
  ans = n - freq_e_sorted[0][1] - freq_o_sorted[1][1]
else:
  ans = n - freq_e_sorted[1][1] - freq_o_sorted[0][1]
print(ans)