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>
#define int long long
using namespace std;
const int INF = 1e10;
const int MOD = 1e9 + 7;
#define dump(x) cout << #x << " = " << (x) << endl;
#define YES(n) cout << ((n) ? "YES" : "NO") << endl
#define Yes(n) cout << ((n) ? "Yes" : "No") << endl
#define SANKOU(n, a, b) cout << ((n) ? (#a) : (#b)) << endl
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define REP(i, n) for (int i = 0; i < (n); ++i)
#define P pair<int, int>
#define V vector<int>
#define M map<int, int>
#define S set<int>
int n, v[100010];
int c1[100010];
int c2[100010];
int m1, m2, n1, n2, v1, v2;

signed main()
{
    cin >> n;
    FOR(i, 0, n)
    {
        cin >> v[i];
        if (i % 2) {
            c1[v[i]]++;
        } else {
            c2[v[i]]++;
        }
    }
    FOR(i, 0, 100010)
    {
        if (m2 < c2[i]) {
            n2 = m2;
            m2 = c2[i];
            v2 = i;
        } else {
            n2 = max(n2, c2[i]);
        }
        if (m1 < c1[i]) {
            n1 = m1;
            m1 = c1[i];
            v1 = i;
        } else {
            n1 = max(n1, c1[i]);
        }
    }
    if (v1 != v2)
        cout << n - m1 - m2 << endl;
    else {
        cout << n - max(m1 + n2, m2 + n1) << endl;
    }
    return 0;
}