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>

struct asumikana {
	int kazu;
	int kosuu;
};

int asumisu(const void* x, const void* y) {
	int a = *(const int*)x, b = *(const int*)y;
	return a < b ? -1 : a > b;
}

int mizuhasu(const void* x, const void* y) {
	struct asumikana a = *(const struct asumikana*)x, b = *(const struct asumikana*)y;
	if (a.kosuu != b.kosuu) return a.kosuu > b.kosuu ? -1 : 1;
	return a.kazu < b.kazu ? -1 : a.kazu > b.kazu;
}

int n;
int v[114514];

int guNums[114514], kiNums[114514];

int guCount, kiCount;
struct asumikana gu[114514], ki[114514];

int main(void) {
	int i;

	if (scanf("%d", &n) != 1) return 1;
	if (n % 2 != 0) {
		puts("n wa guusuu zya nai to dame!!!!!");
		return 2;
	}
	for (i = 0; i < n; i++) {
		if (scanf("%d", &v[i]) != 1) return 1;
	}

	for (i = 1; i < n; i += 2) {
		guNums[i / 2] = v[i - 1];
		kiNums[i / 2] = v[i];
	}
	qsort(guNums, n / 2, sizeof(*guNums), asumisu);
	qsort(kiNums, n / 2, sizeof(*kiNums), asumisu);

	gu[0].kazu = guNums[0];
	gu[0].kosuu = 1;
	guCount = 1;
	for (i = 1; i < n / 2; i++) {
		if (guNums[i] == gu[guCount - 1].kazu) {
			gu[guCount - 1].kosuu++;
		} else {
			gu[guCount].kazu = guNums[i];
			gu[guCount].kosuu = 1;
			guCount++;
		}
	}
	ki[0].kazu = kiNums[0];
	ki[0].kosuu = 1;
	kiCount = 1;
	for (i = 1; i < n / 2; i++) {
		if (kiNums[i] == ki[kiCount - 1].kazu) {
			ki[kiCount - 1].kosuu++;
		} else {
			ki[kiCount].kazu = kiNums[i];
			ki[kiCount].kosuu = 1;
			kiCount++;
		}
	}
	qsort(gu, guCount, sizeof(*gu), mizuhasu);
	qsort(ki, kiCount, sizeof(*ki), mizuhasu);

	if (gu[0].kazu != ki[0].kazu) {
		printf("%d\n", n - gu[0].kosuu - ki[0].kosuu);
	} else if (guCount == 1 && kiCount == 1) {
		printf("%d\n", n / 2);
	} else if (guCount == 1) {
		printf("%d\n", n / 2 - ki[0].kosuu);
	} else if (kiCount == 1) {
		printf("%d\n", n / 2 - gu[0].kosuu);
	} else if (gu[1].kosuu + ki[0].kosuu >= gu[0].kosuu + ki[1].kosuu) {
		printf("%d\n", n - gu[1].kosuu - ki[0].kosuu);
	} else {
		printf("%d\n", n - gu[0].kosuu - ki[1].kosuu);
	}

	return 0;
}