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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>
#include <inttypes.h>
#include <string.h>
#include <math.h>

int main (int argc, char const *argv[])
{
    int64_t n, v;
    int64_t oddnum[(int)1e5] = {0}, evennum[(int)1e5] = {0};
    uint64_t ans = 0, oddmax = 0, evenmax = 0, oddmax2 = 0, evenmax2 = 0, oddi = 0, eveni = 0;

    scanf("%"PRId64, &n);
    for (int i = 0; i < n; i++) {
        scanf("%"PRId64, &v);
        if (i % 2) {
            oddnum[v]++;
        } else {
            evennum[v]++;
        }
    }

    for (int i = 0; i < (int)1e5; i++) {
        if (oddmax < oddnum[i]) {
            oddmax2 = oddmax;
            oddmax = oddnum[i];
            oddi = i;
        } else if (oddmax2 < oddnum[i]) {
            oddmax2 = oddnum[i];
        }
        if (evenmax < evennum[i]) {
            evenmax2 = evenmax;
            evenmax = evennum[i];
            eveni = i;
        } else if (evenmax2 < evennum[i]) {
            evenmax2 = evennum[i];
        }
    }

    if (oddi != eveni) {
        ans = ((n / 2) - oddmax) + ((n / 2) - evenmax);
    } else {
        if (oddmax > evenmax) {
            ans = ((n / 2) - oddmax) + ((n / 2) - evenmax2);
        } else {
            ans = ((n / 2) - oddmax2) + ((n / 2) - evenmax);
        }
    }

    printf("%"PRId64"\n", ans);
}