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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#define rep(i, l, r) for (int i = l; i < r; i++)
#define max(i, j) i > j ? i : j
#define MAX 100001

int main(void) {
  int n;
  scanf("%d", &n);
  int v[MAX];
  rep(i, 0, n)
    scanf("%d", &v[i]);
  
  /* 最頻値を数える */
  int mode1[MAX] = {0};
  int mode2[MAX] = {0};
  // 奇数列
  for (int i = 0; i < n; i += 2)
    mode1[v[i]]++;
  // 偶数列
  for (int i = 1; i < n; i += 2)
    mode2[v[i]]++;

  int max1[2] = {0}, next1 = 0;  // max[頻度][数], 2番目の頻値
  int max2[2] = {0}, next2 = 0; 
  rep(i, 0, MAX) {
    if (mode1[i] > max1[0]) {
      next1 = max1[0];
      max1[0] = mode1[i];
      max1[1] = i;
    } else if (mode1[i] > next1) {
      next1 = mode1[i];
    }
  }
  rep(i, 0, MAX) {
    if (mode2[i] > max2[0]) {
      next2 = max2[0];
      max2[0] = mode2[i];
      max2[1] = i;
    } else if (mode2[i] > next2) {
      next2 = mode2[i];
    }
  }
  
  int ans;
  if (max1[1] == max2[1])
    ans = n - max(next1 + max2[0], max1[0] + next2);
  else
    ans = n - max1[0] - max2[0];
  printf("%d\n", ans);
  return 0;
}