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: Wrong Answer)

#include <stdio.h>
#include<stdlib.h>
int compare(const void *a,const void *b){
    return *(int*)a-*(int*)b;
}
int Min(int a,int b){
    if(a<=b){
        return a;
    }else{
        return b;
    }
}
int main()
{   int ans,nodd,neven,j=0,k=0,i,n,odd[100000],even[100000],oddpr=0,oddse=0,evenpr=0,evense=0,cnt=0;
    scanf("%d",&n);
    for(i=0;i<n;i++){
        if(i%2==0){
            scanf("%d",&odd[j]);
            j++;
        }else{
            scanf("%d",&even[k]);
            k++;
        }
    }
    qsort(odd,j,sizeof(int),compare);
    qsort(even,k,sizeof(int),compare);
    /*for(i=0;i<j;i++){
        printf("%d ",odd[i]);
    }
    for(i=0;i<k;i++){
        printf("%d ",even[i]);
    }
   */
    for(i=0;i<j-1;i++){
        if(odd[i]==odd[i+1]){
            cnt++;
        }else{
            if(cnt>oddpr){
                nodd=odd[i];
                oddse=oddpr;
                oddpr=cnt;
            }else if(cnt>oddse){
                oddse=cnt;
            }
            cnt=0;
        }
    }
    if(cnt>oddpr){
                nodd=odd[i];
                oddse=oddpr;
                oddpr=cnt;
            }else if(cnt>oddse){
                oddse=cnt;
            }
    cnt=0;
    for(i=0;i<k-1;i++){
        if(even[i]==even[i+1]){
            cnt++;
        }else{
            if(cnt>evenpr){
                neven=even[i];
                evense=evenpr;
                evenpr=cnt;
            }else if(cnt>evense){
                evense=cnt;
            }
            cnt=0;
        }
    }
    if(cnt>evenpr){
                neven=even[i];
                evense=evenpr;
                evenpr=cnt;
            }else if(cnt>evense){
                evense=cnt;
            }
    if(neven==nodd&&(evense!=0||oddse!=0)){
        ans=Min(oddpr+evense,evenpr+oddse);
    }else{
        ans=oddpr+evenpr;
    }
    printf("%d",ans);
    
    return 0;
}