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;
#define rep(i,n) for (int i=0; i < (int)(n); i++)
#define rep2(i, s, n) for (int i = (s); i < (int)(n); i++)
using ll = long long;
using VI = vector<int>;
using VVI = vector<vector<int>>;
using P = pair<int, int>;

int main(){
	int n;
    cin >> n;
    VI e(n/2), o(n/2);
    rep(i,n){
        if(i%2 == 0) cin >> e[i/2];
        if(i%2 == 1) cin >> o[i/2];
    }

    sort(e.begin(), e.end());
    sort(o.begin(), o.end());

    vector<P> el(1, P(0,-1));
    vector<P> ol(1, P(0,-1));

    rep(i,n/2){
        if(el.empty() || el.back().second != e[i]){
            P p;
            p.first = 1;
            p.second = e[i];
            el.push_back(p);
        }else{
            el.back().first++;
        }

        if(ol.empty() || ol.back().second != o[i]){
            P p;
            p.first = 1;
            p.second = o[i];
            ol.push_back(p);
        }else{
            ol.back().first++;
        }
    }

    sort(el.begin(), el.end(), greater<P>());
    sort(ol.begin(), ol.end(), greater<P>());

    int ans=0;
    if(el[0].second != ol[0].second){
        ans = n - (el[0].first + ol[0].first);
    }else if(el[0].first > ol[0].first){
        ans = n - (el[0].first + ol[1].first);
    }else{
        ans = n - (el[1].first + ol[0].first);
    }

    cout << ans << endl;

	return 0;
}