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>

int main(void) {
    int n,v[100000],count1[100001],count2[100001],i,max1 = 0,max2 = 0,cmax1,cmax2,nmax1 = 0,nmax2 = 0;
    scanf("%d",&n);
    for (i = 0;i < n;i++) {
        scanf("%d",&v[i]);
    }
    for (i = 0;i <= 100000;i++) {
        count1[i] = 0;
        count2[i] = 0;
    }
    for (i = 0;i < n;i++) {
        if (i % 2 == 0) {
            count1[v[i]]++;
        } else {
            count2[v[i]]++;
        }
    }
    for (i = 1;i <= 100000;i++) {
        if (count1[i] > max1) {
            max1 = count1[i];
            cmax1 = i;
        }
        if (count2[i] > max2) {
            max2 = count2[i];
            cmax2 = i;
        }
    }
    if (cmax1 == cmax2) {
        count1[cmax1] = 0;
        count2[cmax2] = 0;
        for (i = 1;i <= 100000;i++) {
            if (count1[i] > nmax1) {
                nmax1 = count1[i];
                cmax1 = i;
            }
            if (count2[i] > nmax2) {
                nmax2 = count2[i];
                cmax2 = i;
            }
        }
        if (max1 + nmax2 > max2 + nmax1) {
            printf("%d\n",n - (max1 + nmax2));
        } else {
            printf("%d\n",n - (max2 + nmax1));
        }
    } else {
        printf("%d\n",n - (max1 + max2));
    }
    return 0;
}