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;

bool cmp(pair<int, int> x, pair<int, int> y) {
	return x.second > y.second;
}

int n;

int main(){
    ios_base::sync_with_stdio(false); //Fast I/O
    cin.tie(NULL);

    cin >> n;
    map<int, int> a, b;
	vector<int> v1, v2;
	vector<pair<int, int>> va, vb;

	for(int i=0;i<n;i++) {
		int t;
		cin >> t;
		if(i % 2) {
			if(b[t] == 0)
				v2.push_back(t);
			b[t]++;
		} else {
			if(a[t] == 0)
				v1.push_back(t);
			a[t]++;
		}
	}
	for(int x:v1) {
		va.push_back({x, a[x]});
	}
	for(int x:v2) {
		vb.push_back({x, b[x]});
	}
	int k = (n+1)/2, m = n/2;
	sort(va.begin(), va.end(), cmp);
	sort(vb.begin(), vb.end(), cmp);
	int v = k - va[0].second,
	    u = m - vb[0].second;
	int ans = u + v;
	if(va[0].first == vb[0].first) {
		int u1, v1;
		if(va.size()>1)
			u1 = k - va[1].second;
		else
			u1 = k;
		if(vb.size()>1)
			v1 = m - vb[1].second;
		else
			v1 = m;
		ans = min(v+v1, u+u1);
	}
	cout << ans << endl;

    return 0;
}