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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <climits>
#include <vector>
#include <map>
#include <set>
#include <list>
#include <stack>
#include <queue>
#include <algorithm>
#include <iostream>
#include <string>

#define REP(i,n) for(auto i=0;i<n;++i)
#define REPR(i,n) for(auto i=n;i>=0;--i)
#define REPI(itr,v) for(auto itr=v.begin();itr!=v.end();++itr)
#define REPIR(itr,v) for(auto itr=v.rbegin();itr!=v.rend();++itr)
#define FOR(i,a,b) for(auto i=a;i<b;++i)
#define SORT(v,n) sort(v, v+n)
#define SORTV(v) sort(v.begin(), v.end())
#define ALL(v) v.begin(),v.end()
#define llong long long
#define ll long long
#define INF 999999999
#define MOD 1000000007
#define pb push_back
#define pf push_front
#define MP make_pair
#define SV(n,v) {int tmp;for(int i=0;i<n;++i){scanf("%d",&tmp);v.push_back(tmp);}}

int dx[] = {0, 0, -1, 1};
int dy[] = {1, -1, 0, 0};

using namespace std;

typedef pair<int,int> pii;

int main(){

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

  int even[100001] = {}, odd[100001] = {};
  REP(i,n){
    int v;
    scanf("%d", &v);
    if(i % 2 == 0){
      odd[v] += 1;
    }else{
      even[v] += 1;
    }
  } 

  int odd1 = 0, even1 = 0, odd2, even2;
  int odd1_idx, even1_idx;
  REP(i,100001){
    if(odd[i] >= odd1){
      odd2 = odd1;
      odd1 = odd[i];
      odd1_idx = i;
    }
    if(even[i] >= even1){
      even2 =  even[i];
      even1 = even[i];
      even1_idx = i;
    }
  }
  if(odd1 == n / 2 && odd1_idx == even1_idx){
    printf("%d\n", n - even1);
  }else if(even1 == n / 2 && odd1_idx == even1_idx){
    printf("%d\n", n - odd1);
  }else if(odd1_idx != even1_idx){
    printf("%d\n", n - odd1 - even1);
  }else{
    printf("%d\n", min(n-odd1-even2, n-odd2-even1));
  }

  return 0;
}