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 endl '\n'
#define PB push_back
#define MP make_pair
#define LL long long
#define FOR(i,a,b) for(int i = (a); i <= (b); i++)
#define RE(i,n) FOR(i,1,n)
#define REP(i,n) FOR(i,0,(int)(n)-1)
#define R(i,n) REP(i,n)
#define VI vector<int>
#define PII pair<int,int>
#define LD long double
#define FI first
#define SE second
#define st FI
#define nd SE
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)(x).size())

typedef long long ll;

bool file=0,rt=1;
clock_t tStart ;
void FAST_IO();
////////////////////////

vector<int>  arr(100005);
int a[100005];
int main() {

    //FAST_IO();
    ios_base::sync_with_stdio(0);
    //cin.tie(NULL);
    cout << setprecision(5);
    cout << fixed;

    int n;
    cin >> n;





    for(int i = 0; i < n; i++) {

        cin >> a[i];

    }

    for(int i = 0; i < n; i++) {

        arr[a[i]]++;
    }

    int val = max_element(arr.begin(), arr.end()) - arr.begin();
    int pos = find(arr.begin(), arr.end(), val) - arr.begin();


    int mx = 0;
    for(int i = 0; i < n; i++) {

        if(a[i] == val) continue;

        else if(a[i] > mx) {

            mx = i;
        }
    }

    int ans1 = 0; int ans2 = 0;

    for(int i = pos; i < n; i+=2) {

        if(pos+2 < n) {

            if(a[pos] != a[pos+2]) {

                a[pos+2] = a[pos];
                ans1++;
            }
        }
    }


    for(int i = pos; i >= 0; i-=2) {

        if(pos-2 >= 0) {

            if(a[pos] != a[pos-2]) {

                a[pos-2] = a[pos];
                ans1++;
            }
        }
    }

    pos = mx;

    for(int i = pos; i < n; i+=2) {

        if(pos+2 < n) {

            if(a[pos] != a[pos+2]) {

                a[pos+2] = a[pos];
                ans2++;
            }
        }
    }


    for(int i = pos; i >= 0; i-=2) {

        if(pos-2 >= 0) {

            if(a[pos] != a[pos-2]) {

                a[pos-2] = a[pos];
                ans2++;
            }
        }
    }


//    if(a[0] == a[1]) {
//
//
//
//        a[1] = a[0] + 1;
//
//        for(int i = 1; i < n; i+=2) {
//
//            if(i+2 < n) {
//
//                if(a[i] != a[i+2]) {
//
//                    a[i+2] = a[i];
//                    ans1++;
//                }
//            }
//
//        }
//        ans1++;
//    }

    if(a[0] == a[1]) {

        cout << n/2 << endl;
        return 0;
    }

    cout << ans1 + ans2 << endl;






    return 0;
}

void FAST_IO()
{
    ios_base::sync_with_stdio(0);
    cin.tie(NULL);
    //cout.tie(NULL);
    //cout.setf(ios::fixed);
    //cout.precision(2);
    if(rt && file){ tStart = clock(); }
    if(file){
    #ifndef _offline
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    #endif
    }
}