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 <iostream>
#include <set>
#include <vector>
#include <string>
#include <algorithm>
#include <iterator>
#include <map>
#include <utility>
#include <list>
#include <climits>
#include <bitset>
#include <numeric>

using namespace std;

#define rep(i, n) for(int i = 0; i < n; i++)
#define repi(i, n) for(auto i = (n).begin(); i != (n).end(); i++)
#define in_arr(type, a, n) copy_n(istream_iterator<type>(cin), n, (a).begin());
#define ll long long int

int main(){
    int n;
    cin >> n;
    vector<int> a(n);
    in_arr(int, a, n);
    map<int, int> m;
    for(int i = 0; i < n; i += 2){
        ++m[a[i]];
    }
    int mv = 0, mi;
    repi(i, m){
        if(mv < i->second){
            mv = i->second;
            mi = i->first;
        }
    }
    int ans = 0;
    for(int i = 0; i < n; i += 2){
        if(a[i] != mi){
            ++ans;
        }
    }
    //
    m.clear();
    int prev_mi = mi;
    for(int i = 1; i < n; i += 2){
        if(prev_mi == a[i])
            continue;
        ++m[a[i]];
    }
    mv = 0;
    mi = INT_MAX;
    repi(i, m){
        if(mv < i->second){
            mv = i->second;
            mi = i->first;
        }
    }
    for(int i = 1; i < n; i += 2){
        if(a[i] != mi){
            ++ans;
        }
    }
    if(mi == prev_mi && m.size() == 0){
        ans += n / 2;
    }
    cout << ans << endl;
    return 0;
}