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 <vector>
#include <utility>
#include <unordered_map>
using namespace std;

int main(){
    int n;
    cin >> n;
    vector<int> v(n);
    unordered_map <int, int> odd;
    unordered_map <int, int> even;
    int max_odd = 0, max_even = 0, sec_odd = 0, sec_even = 0;
    int odd_num, even_num;
    for (int i;i<n;i++){
        cin >> v[i];
        if (i % 2){
            if (odd.find(v[i]) == odd.end()){
                odd.insert(make_pair(v[i], 1));
                if (odd[v[i]] > max_odd){
                    max_odd = odd[v[i]];
                    odd_num = v[i];
                }
                else if(odd[v[i]] > sec_odd){
                    sec_odd = v[i];
                }
            }

            else{
                odd[v[i]]++;
                if (odd[v[i]] > max_odd){
                    if (v[i] != odd_num)
                        sec_odd = max_odd;
                    max_odd = odd[v[i]];
                    odd_num = v[i];
                }
            }

        }
        else{
            if (even.find(v[i]) == even.end()){
                even.insert(make_pair(v[i], 1));
                if (even[v[i]] > max_even){
                    max_even = even[v[i]];
                    even_num = v[i];
                }
                else if(even[v[i]] > sec_even){
                    sec_even = v[i];
                }
            }

            else{
                even[v[i]]++;
                if (even[v[i]] > max_even){
                    if (v[i] != even_num)
                        sec_even = max_even;
                    max_even = even[v[i]];
                    even_num = v[i];
                }
            }
        }
    }
    int ret = n - max_odd - max_even;
    if (odd_num == even_num){
        if ((max_even + sec_odd) > (max_odd + sec_even))
            ret = n - max_even - sec_odd;
        else
            ret = n - max_odd - sec_even;
    }

    cout << ret;
    return 0;

}