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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
#define FOR(i,l,r) for(long long i=(l);i<(r);++i)
#define REP(i,n) FOR(i,0,n)
#define REPS(i,n) FOR(i,1,n+1)
#define int long long
#define pb push_back
#define eb emplace_back
#define mp make_pair
#define mt make_tuple
const int MOD=1e9+7;
const int INF=1e15;
 
signed main(){
    int N;cin>>N;
    map<int,int>amemo;
    map<int,int>bmemo;
    REP(i,N){
        int v;cin>>v;
        if(i%2==0)amemo[v]++;
        else bmemo[v]++;
    }
    int max1=0;
    int max1v=0;
    int max2=0;
    int max2v=0;
    for(auto it=amemo.begin();it!=amemo.end();it++){
        int d=it->second;
        if(max1<d){
            max1=d;
            max1v=it->first;
        }
        else{
            if(max2<d){
                max2=d;
                max2v=it->first;
            }
        }
    }
    int bmax1=0;
    int bmax1v=0;
    int bmax2=0;
    int bmax2v=0;
    for(auto it=bmemo.begin();it!=bmemo.end();it++){
        int d=it->second;
        if(bmax1<d){
            bmax1=d;
            bmax1v=it->first;
        }
        else{
            if(bmax2<d){
                bmax2=d;
                bmax2v=it->first;
            }
        }
    }
  	if(max1v!=bmax1v)cout<<N-max1-bmax1<<endl;
    else cout<<N-max1-max(max2,bmax2)<<endl;
}