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<bits/stdc++.h>
#pragma warning(disable:4996)
using namespace std;
using ll = long long;
using ld = long double;
const int dx[4] = { 1,0,-1,0 }, dy[4] = { 0,1,0,-1 };
//const long double PI = acos(-1.0);
#define int long long
signed main() {
	int n;
	cin >> n;
	vector<int>v(n);
	unordered_map<int, int>ump0;
	unordered_map<int, int>ump1;
	for (int i = 0; i < n; i++) {
		cin >> v[i];
		if (i % 2 == 0) {
			ump0[v[i]]++;
		}
		else if (i % 2 == 1) {
			ump1[v[i]]++;
		}

	}

	pair<int, int> f0 = { 0,0 };
	pair<int, int> s0 = { 0,0 };
	for (auto e : ump0) {
		int key, value;
		tie(key, value) = e;
		if (f0.second < value) {
			
			f0 = e;
		}
		else if (value <= f0.second && s0.second<value) {
			s0 = e;
		}
	}
	pair<int, int> f1 = { 0,0 };
	pair<int, int> s1 = { 0,0 };
	for (auto e : ump1) {
		int key, value;
		tie(key, value) = e;
		if (f1.second < value) {

			f1 = e;
		}
		else if (value <= f1.second && s1.second<value) {
			s1 = e;
		}
		
	}
	if (ump0.size() == 1 && ump1.size() == 1 && ump0 == ump1) {
		cout << n / 2 << endl;
	}
	else if (f0.first == f1.first) {
		cout << min(n - f0.second - s1.second, n - f1.second - s0.second) << endl;
	}
	else {
		cout << n - f0.second - f1.second << endl;
	}
	return 0;
}