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 <iostream>
#include <math.h>
#include <algorithm>
#include <set>
using namespace std;
typedef long long ll;
#define rep(i,s,n)for(ll i=s;i<n;i++)
#define repe(i,s,n)for(ll i=s;i<=n;i++)
#define rep_r(i,s,n)for(ll i=n-1;i>=s;i--)
#define repe_r(i,s,n)for(ll i=s;i<=n;i++)
static const ll MOD = 1e9 + 7;
static const ll MAX_L = (ll)1 << 62;

ll a[100001] = {};
ll b[100001] = {};
ll A[100001] = {};
ll B[100001] = {};
set<ll> AS;
set<ll> BS;
int main() {
	ll n; cin >> n;
	repe(i, 1, n / 2) {
		ll x, y; cin >> x >> y;
		AS.insert(x);
		BS.insert(y);
		a[x]++, A[x]++, b[y]++, B[y]++;
	}

	sort(A + 1, A + 100001);
	reverse(A + 1, A + 100001);
	sort(B + 1, B + 100001);
	reverse(B + 1, B + 100001);
	sort(a + 1, a + 100001);
	reverse(a + 1, a + 100001);
	sort(b + 1, b + 100001);
	reverse(b + 1, b + 100001);

	if (AS.size() == 1 && BS.size() == 1) {
		if (*AS.begin() == *BS.begin()) {
			cout << n / 2 << endl;
		}
		else {
			cout << 0 << endl;
		}
		return 0;
	}
	////////////////////////////////////////////////////////////////
	if (AS.size() == 1) {
		cout << n - A[1] - B[2] << endl;
	}
	else if (BS.size() == 1) {
		cout << n - B[1] - A[2] << endl;
	}
	else {
		if (A[1] != B[1]) {
			cout << n - A[1] - B[1] << endl;
			return 0;
		}
		else {
			ll ans1 = n - A[1] - B[2];
			ll ans2 = n - A[2] - B[1];
			cout << min(ans1, ans2) << endl;
			return 0;
		}
	}

	return 0;
}