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>

typedef struct Number
{
    int value;
    int changes;
} Number;

#define MAX_VALUE 100000

void GetBest(int *count, int total, Number *first, Number *second)
{
    *first = (Number){.value = 1, .changes = total - count[1]};
    *second = (Number){.value = -1, .changes = total + 1};

    for (int i = 2; i <= MAX_VALUE; i += 1) {
        Number new_num = {.value = i, .changes = total - count[i]};
        if (new_num.changes < first->changes) {
            *second = *first;
            *first = new_num;
        } else if (new_num.changes < second->changes) {
            *second = new_num;
        }
    }
}

int Min(int a, int b) { return a < b ? a : b; }

int count[2][MAX_VALUE + 1];

int main()
{
    int n;
    scanf("%d", &n);

    for (int i = 0; i < n; i += 1) {
        int num;
        scanf("%d", &num);
        count[i % 2][num] += 1;
    }

    Number even1, even2;
    GetBest(count[0], (n + 1) / 2, &even1, &even2);

    Number odd1, odd2;
    GetBest(count[1], n / 2, &odd1, &odd2);

    int res = even1.changes + odd1.changes;
    if (even1.value == odd1.value) {
        res = Min(even2.changes + odd1.changes,
                  even1.changes + odd2.changes);
    }

    printf("%d\n", res);
    return 0;
}