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 <iostream>
#include <stdio.h>
#include <algorithm>
#include <string>
#define maxn 100005
using namespace std;

int T[maxn << 2][2];
int v[maxn][2];
int a[maxn];

void update(int rt, int l, int r, int pos, int f) {
	if (l == r) {
		T[rt][f] += 1;
		return;
	}
	int mid = (l + r) >> 1;
	if (mid >= pos) update(rt << 1, l, mid, pos, f);
	else	update(rt << 1 | 1, mid + 1, r, pos, f);
	T[rt][f] = max(T[rt << 1][f], T[rt << 1 | 1][f]);
}

int query(int rt, int l, int r, int L, int R, int f) {
	if (L <= l && R >= r) return T[rt][f];
	int mid = (l + r) >> 1, mx = 0;
	if (mid >= L)	mx = max(mx, query(rt << 1, l, mid, L, R, f));
	if (mid < R)	mx = max(mx, query(rt << 1 | 1, mid + 1, r, L, R, f));
	return mx;
}

int main() {
	int n;	scanf("%d", &n);
	for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
	for (int i = 1; i <= n; i += 2) update(1, 1, maxn - 1, a[i], 1);
	for (int i = 1; i <= n; i += 2) v[a[i]][1]++;
	for (int i = 2; i <= n; i += 2) update(1, 1, maxn - 1, a[i], 0);
	for (int i = 2; i <= n; i += 2) v[a[i]][0]++;
	int mi = 0x3f3f3f3f;
	for (int i = 1; i <= n; i += 2) {
		int tmp_sum = v[a[i]][1] + query(1, 1, maxn - 1, a[i] + 1, maxn - 1, 0);
		mi = min(mi, n - tmp_sum);
	}
	for (int i = 2; i <= n; i += 2) {
		int tmp_sum = v[a[i]][0] + query(1, 1, maxn - 1, a[i] + 1, maxn - 1, 1);
		mi = min(mi, n - tmp_sum);
	}
	printf("%d\n", mi);
	return 0;
}