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;
const int maxx = 1e5 + 7;
int n;
int ans, res;
int a1, a2;
int a[maxx];
map <int, int> c1;
map <int, int> c2;

int main() {
	while(scanf("%d", &n) != EOF) {
		c1.clear();
		c2.clear();
		ans = 0, res = 0;
		for(int i = 1; i <= n; i++) {
			scanf("%d", &a[i]);
			if(i & 1) {
				c1[a[i]]++;
				if(ans < c1[a[i]]) {
					a1 = a[i];
					ans = c1[a[i]];
				}
			}
			else {
				c2[a[i]]++;
				if(res < c2[a[i]]) {
					a2 = a[i];
					res = c2[a[i]];
				}
			}
		}
		if(a1 == a2){
			if(ans == res) {
				int sum1 = 0, sum2 = 0;
				for(int i = 1; i <= n; i++) {
					if(a1 == a[i]) continue;
					if(i & 1) sum1 = max(sum1, c1[a[i]]);
					else sum2 = max(sum2, c2[a[i]]);
				}
				if(sum1 >= sum2) ans = sum1;
				else res = sum2;
			}
			else if(ans < res) {
				ans = 0;
				for(int i = 1; i <= n; i += 2){
					if(a[i] != a2) ans = max(ans, c1[a[i]]);
				}
			}
			else {
				res = 0;
				for(int i = 2; i <= n; i += 2){
					if(a[i] != a1) res = max(res, c2[a[i]]);
				}
			}
		}
		printf("%d\n", n - ans - res);
	}
}