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: Accepted)

#include <bits/stdc++.h>
using namespace std;
using pii = pair<int, int>;

int main() {
    int N; scanf("%d",&N);
    vector<int> a, b;
    for(int i=0; i<N; i++) {
        int foo; scanf("%d",&foo);
        if(i%2) a.push_back(foo);
        else b.push_back(foo);
    }
    sort(a.begin(), a.end());
    sort(b.begin(), b.end());

    vector<pii> as, bs;
    {
        int c, p = -1;
        for(auto& it: a) {
            if(p != it) {
                if(p != -1) as.push_back({c, p});
                p = it;
                c = 1;
            }
            else ++c;
        }
        as.push_back({c, p});
    }
    {
        int c, p = -1;
        for(auto& it: b) {
            if(p != it) {
                if(p != -1) bs.push_back({c, p});
                p = it;
                c = 1;
            }
            else ++c;
        }
        bs.push_back({c, p});
    }

    sort(as.begin(), as.end()); reverse(as.begin(), as.end());
    sort(bs.begin(), bs.end()); reverse(bs.begin(), bs.end());
    as.push_back({0, 1e9}); bs.push_back({0, 1e9});

    if(as[0].second == bs[0].second)
        printf("%d", min(N - as[0].first - bs[1].first, N - as[1].first - bs[0].first));
    else
        printf("%d", N - as[0].first - bs[0].first);
    return 0;
}