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>

#define fto(i, a, b) for (int i = a; i <= b; ++i)
#define fdto(i, a, b) for (int i = a; i >= b; --i)
#define ff first
#define ss second
#define pii pair<int, int>
#define pb push_back
#define mp make_pair
#define ll long long
#define pi = 3.1415926535897
#define oo 1000000007
#define OO 1000000000000000007

using namespace std;

bool cmp (pii a, pii b) {
    return a.ff < b.ff;
}

int n;
int f[100005];
int g[100005];
int a[100005];
vector <pii> even;
vector <pii> odd;

int main() {
//    freopen("codef.inp", "r", stdin);
//    freopen("codef.out", "w", stdout);

    scanf("%d", &n);
    fto(i, 1, n) scanf("%d", &a[i]);

    fto(i, 1, n) {
        if (i%2) {
            if (!f[a[i]]) odd.pb(mp(1, a[i]));
            ++f[a[i]];
        }
        else {
            if (!g[a[i]]) even.pb(mp(1, a[i]));
            ++g[a[i]];

        }
    }

    if (odd.size() == 1 && even.size() == 1) {
        if (a[1] == a[2]) printf("%d\n", n/2);
        if (a[1] != a[2]) printf("%d\n", 0);
    }
    else {

        fto(i, 0, (int) odd.size()-1) {
            odd[i].ff = f[odd[i].ss];
        }

        fto(i, 0, (int) even.size()-1) {
            even[i].ff = g[even[i].ss];
        }
        sort(odd.begin(), odd.end(), cmp);
        sort(even.begin(), even.end(), cmp);

        int res;

        if (odd.size() == 1) {
            if (odd[0].ss == even[0].ss) res = n/2-even[1].ff;
            else res = n/2-even[0].ff;
        }

        if (even.size() == 1) {
            if (odd[0].ss == even[0].ss) res = n/2-odd[1].ff;
            else res = n/2-odd[0].ff;
        }

        if (even.size() != 1 && odd.size() != 1) {
            res = n-odd[0].ff-even[0].ff;
            if (odd[0].ss == even[0].ss)
                n = min(n-odd[0].ss-even[1].ss, n-odd[1].ss-even[0].ss);
        }

        printf("%d\n", res);
    }


    return 0;
}