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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
int max(int x, int y) {
	return x > y ? x : y;
}
int cmpint(const void *x, const void *y) {
	return *(int *)y - *(int *)x;
}
int main() {
	int n, v, EvenNumCnt[100000] = {0}, OddNumCnt[100000] = {0};
	int OddMax = 0, EvenMax = 0;
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		scanf("%d", &v);
		if (i % 2 == 1) {
			OddNumCnt[v - 1]++;
		} else {
			EvenNumCnt[v - 1]++;
		}
	}
	int OddMaxNum, EvenMaxNum;
	for (int i = 0; i < 1e5; i++) {
		if (OddMax < OddNumCnt[i]) {
			OddMax = OddNumCnt[i];
			OddMaxNum = i + 1;
		}
	}
	for (int i = 0; i < 1e5; i++) {
		if (EvenMax < EvenNumCnt[i]) {
			EvenMax = EvenNumCnt[i];
			EvenMaxNum = i + 1;
		}
	}
	if (OddMaxNum == EvenMaxNum) {
		qsort(OddNumCnt, n, sizeof(int), cmpint);
		qsort(EvenNumCnt, n, sizeof(int), cmpint);
		OddMax = max(OddNumCnt[1], EvenNumCnt[1]);
	}
	printf("%d\n", n / 2 - OddMax + n / 2 - EvenMax);
	return 0;
}