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;

// clang-format off
#define  range(i, l, r) for (int i = (int)(l); i < (int)(r); ++(i))
#define rrange(i, l, r) for (int i = (int)(r) - 1; i >= (int)(l); --(i))

#define  whole(f, x, ...) ([&](decltype((x)) container) { return (f)(  begin(container),  end(container), ## __VA_ARGS__); })(x)
#define rwhole(f, x, ...) ([&](decltype((x)) container) { return (f)( rbegin(container), rend(container), ## __VA_ARGS__); })(x)

#define endl '\n'
#define debug(x) cerr << "(" << __LINE__ << ")" << #x << ": " << (x) << endl

using i32 = int;
using u32 = unsigned int;
using i64 = long long;
using u64 = unsigned long long;

constexpr i32 mod   = 1e9 + 7;
constexpr i32 inf   = 1001001001;
constexpr i64 infll = 1001001001001001001ll;

constexpr int dx[] = {0, -1, 1, 0, -1, 1, -1, 1}; 
constexpr int dy[] = {-1, 0, 0, 1, -1, -1, 1, 1};

struct IoSetup { IoSetup(int x = 15){ cin.tie(0); ios::sync_with_stdio(0); cout << fixed << setprecision(x); cerr << fixed << setprecision(x); } } iosetup;

template <typename T = i64> T input() { T x; cin >> x; return x; }

template <typename T> ostream &operator<<(ostream &os, vector<T> &v) { range(i, 0, v.size()) { os << v[i] << (i + 1 != v.size() ? " " : ""); } return os; } 
template <typename T> istream &operator>>(istream &is, vector<T> &v) { for (T &in : v) is >> in; return is; }

template <typename T1, typename T2> ostream &operator<<(ostream &os, pair<T1, T2> p) { os << "(" << p.first << ", " << p.second << ")"; return os; }
template <typename T1, typename T2> istream &operator>>(istream &is, pair<T1, T2> &p) { is >> p.first >> p.second; return is; }

template <typename T> vector<T> make_vector(size_t a, T b) { return vector<T>(a, b); }
template <typename... Ts> auto make_vector(size_t a, Ts... ts) { return vector<decltype(make_vector(ts...))>(a, make_vector(ts...)); }

template <typename T1, typename T2> inline bool chmax(T1 &a, T2 b) { return a < b && (a = b, true); }
template <typename T1, typename T2> inline bool chmin(T1 &a, T2 b) { return a > b && (a = b, true); }
// clang-format on

void solver() {
  i32 n = input();
  auto cnt = make_vector(2, 100001, 0);
  
  range(i, 0, n) {
    cnt[i & 1][input()]++;
  }

  vector< pair<i32, i32> > od, ev;
  range(i, 0, 100001) {
    ev.emplace_back(cnt[0][i], i);
    od.emplace_back(cnt[1][i], i);
  }

  rwhole(sort, ev);
  rwhole(sort, od);

  i32 ans = n;
  if (ev[0].second == od[0].second) {
    ans -= max(ev[0].first + od[1].first, ev[1].first + od[0].first);
  } else {
    ans -= ev[0].first + od[0].first;
  }

  cout << ans << endl;
}

signed main(int argc, char *argv[]) {
  solver();
}