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)

// unolight {{{
#include <bits/stdc++.h>
#include <unistd.h>
#define SZ(x) ((int)(x).size())
#define ALL(c) begin(c),end(c)
#define REP(i,n) for ( int i=0; i<(int)(n); i++ )
#define REP1(i,a,b) for ( int i=(int)(a); i<=(int)(b); i++ )
#define FOR(it,c) for ( auto it=(c).begin(); it!=(c).end(); it++ )
#define MP make_pair
#define PB push_back
using namespace std;
typedef pair<int,int> PII;
typedef vector<int> VI;
typedef long long LL;

void RI() {}

template<typename... T>
void RI( int& head, T&... tail ) {
    scanf("%d",&head);
    RI(tail...);
}

template<typename T>
void _dump( const char* s, T&& head ) { cerr<<s<<"="<<head<<endl; }

template<typename T, typename... Args>
void _dump( const char* s, T&& head, Args&&... tail ) {
    int c=0;
    while ( *s!=',' || c!=0 ) {
        if ( *s=='(' || *s=='[' || *s=='{' ) c++;
        if ( *s==')' || *s==']' || *s=='}' ) c--;
        cerr<<*s++;
    }
    cerr<<"="<<head<<", ";
    _dump(s+1,tail...);
}

#define dump(...) _dump(#__VA_ARGS__, __VA_ARGS__);

template<typename A, typename B>
ostream& operator <<( ostream &s, const pair<A,B> &p ) { return s<<"("<<p.first<<","<<p.second<<")"; }

template<typename Iter>
ostream& _out( ostream &s, Iter b, Iter e ) {
    s<<"[";
    for ( auto it=b; it!=e; it++ ) s<<(it==b?"":" ")<<*it;
    s<<"]";
    return s;
}

template<typename T>
ostream& operator <<( ostream &s, const vector<T> &c ) { return _out(s,ALL(c)); }
template<typename T>
ostream& operator <<( ostream &s, const set<T> &c ) { return _out(s,ALL(c)); }
template<typename A, typename B>
ostream& operator <<( ostream &s, const map<A,B> &c ) { return _out(s,ALL(c)); }

// }}}
const int MAX_N = 100005;
int n;
map<int,int> odd;
map<int,int> even;

int main( int argc, char *argv[] ) {
    RI(n);
    int max_odd[2]={}, max_even[2]={};
    int ov, ev;
    REP(i, n) {
        int a; RI(a);
        if(i%2) {
            odd[a]++;
        } else {
            even[a]++;
        }
    }
    FOR(i, odd) {
        int a = i->first;
        if(odd[a] > max_odd[0]) {
            max_odd[1] = max_odd[0];
            max_odd[0] = odd[a];
            ov = a;
        } else if(odd[a] > max_odd[1]) {
            max_odd[1] = odd[a];
        }
    }
    FOR(i, even) {
        int a = i->first;
        if(even[a] > max_even[0]) {
            max_even[1] = max_even[0];
            max_even[0] = even[a];
            ev = a;
        } else if(even[a] > max_even[1]) {
            max_even[1] = even[a];
        }
    }
    if(ev == ov) {
        if(max_even[0] > max_odd[0]) {
            printf("%d\n", n - max_even[0] - max_odd[1]);
        } else {
            printf("%d\n", n - max_even[1] - max_odd[0]);
        }
    } else {
        printf("%d\n", n - max_even[0] - max_odd[0]);
    }
    return 0;
}