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 <bits/stdc++.h>
using  namespace std;

int main(){
    int n,v,ans,c1,c2;
    cin >> n;
    map<int, int> V0,V1;
    pair<int, int> v1top = make_pair(-1, 0),v1sec = make_pair(-1, 0), v0top = make_pair(-1, 0), v0sec = make_pair(-1, 0);

    
    for(int i=0;i<n;i++){
        cin >> v;
        if(i%2){
            V1[v]++;
        }else{
            V0[v]++;
        }
    }

    for(auto iter = V0.begin();iter!=V0.end();iter++){
        if(v0top.second < iter->second){
            v0top.first = iter->first;
            v0top.second = iter->second;
        }else if(v0sec.second < iter->second){
            v0sec.second = iter->second;
        }
    }
  
    for(auto iter = V1.begin();iter!=V1.end();iter++){
        if(v1top.second < iter->second){
            v1top.first = iter->first;
            v1top.second = iter->second;
        }else if(v1sec.second < iter->second){
            v1sec.second = iter->second;
        }
    }
  
    if(v0top.first==v1top.first){
            c1 = n-v0top.second-v1sec.second;
            c2 = n-v0sec.second-v1top.second;
            ans = min(c1, c2);
    }else{
        ans = n-v0top.second-v1top.second;
    }
    cout << ans << endl;
}