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 <algorithm>
#include <iomanip>
#include <iostream>
#include <set>
#include <string>
#include <vector>
#define SORT(a) sort((a).begin(), (a).end())
#define RSORT(a) sort((a).rbegin(), (a).rend())
#define REP(i, n) for (int i = 0; i < n; i++)
#define FOR(i, start, end) for (int i = start; i < end; i++)
#define FORSTEP(i, start, end, step) for (int i = start; i < end; i += step)
#define ALL(a) a.begin(), a.end()
typedef long long ll;
using namespace std;

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    int N;
    cin >> N;
    vector<int> A(1E6), B(1E6);
    REP(i, N) {
        int x;
        cin >> x;
        if (i % 2 == 0) {
            A[x]++;
        } else {
            B[x]++;
        }
    }

    int amax = 0, bmax = 0;
    int amax2 = 0, bmax2 = 0;
    int ai, bi;
    REP(i, 1E6) {
        if (A[i] < amax && 0 < A[i])
            amax2 = max(amax2, A[i]);
        if (B[i] < bmax && 0 < A[i])
            bmax2 = max(bmax2, B[i]);
        if (amax <= A[i]) {
            ai    = i;
            amax2 = amax;
            amax  = A[i];
        }
        if (bmax <= B[i]) {
            bi    = i;
            bmax2 = bmax;
            bmax  = B[i];
        }
    }
    int ans;
    if (ai != bi) {
        ans = N - amax - bmax;
    } else {
        if (amax2 > bmax2) {
            ans = N - amax2 - bmax;
        } else {
            ans = N - amax - bmax2;
        }
    }

    cout << ans << endl;

    return 0;
}