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: Time Limit Exceeded)

#include <stdio.h>

int main() {
  int n;
  scanf("%d",&n);

  if (n % 2 == 1) {
    printf("Error");
    return 0;
  }

  int o[n/2];
  int e[n/2];
  for (int i = 0; i < n/2; i++) {
    scanf("%d", &o[i]);
    getchar();
    scanf("%d", &e[i]);
    getchar();
  }

  //o,vの最頻値判定
  int temp_cnt = 0;
  int ofst_cnt = 0;
  int oscd_cnt = 0;
  int ofirst = 0;
  int osecond;
  int othird;
  for (int i = 0; i < n/2; i++) {
    temp_cnt = 1;
    if (ofirst == o[i])
      continue;
    if (osecond == o[i])
      continue;
    if (othird == o[i])
      continue;
    for (int j = i+1; j < n/2; j++) {
      if (o[i] == o[j])
        temp_cnt++;
    }
    if (temp_cnt > ofst_cnt) {
      osecond = ofirst;
      oscd_cnt = ofst_cnt;
      ofirst = o[i];
      ofst_cnt = temp_cnt;
    } else if (temp_cnt > oscd_cnt) {
      if (ofirst != o[i]) {
        othird = osecond;
        osecond = o[i];
        oscd_cnt = temp_cnt;
      }
    }
  }

  int efst_cnt = 0;
  int escd_cnt = 0;
  int efirst = 0;
  int esecond;
  int ethird;
  for (int i = 0; i < n/2; i++) {
    temp_cnt = 1;
    if (efirst == e[i])
      continue;
    if (esecond == e[i])
      continue;
    if (ethird == e[i])
      continue;
    for (int j = i+1; j < n/2; j++) {
      if (e[i] == e[j])
        temp_cnt++;
    }
    if (temp_cnt > efst_cnt) {
      esecond = efirst;
      escd_cnt = efst_cnt;
      efirst = e[i];
      efst_cnt = temp_cnt;
    } else if (temp_cnt > escd_cnt) {
      if (efirst != e[i]) {
        ethird = esecond;
        esecond = e[i];
        escd_cnt = temp_cnt;
      }
    }
  }

  int ans;
  if (ofirst != efirst) {
    ans = n - ofst_cnt - efst_cnt;
  } else {
    if (ofst_cnt + escd_cnt < efst_cnt + oscd_cnt) {
      ans = n - efst_cnt - oscd_cnt;
    } else if (ofst_cnt + escd_cnt > efst_cnt + oscd_cnt) {
      ans = n - ofst_cnt - escd_cnt;
    } else {
      ans = n - ofst_cnt - escd_cnt;
    }
  }
  printf("%d\n", ans);
}