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: Accepted)

#include <bits/stdc++.h>
using namespace std;

int main(){
#ifdef LOCAL
    freopen("../a.txt", "r", stdin);
#endif
    ios::sync_with_stdio(0);
    cin.tie(0);
    int n;
    cin >> n;
    vector<int> a(n);
    for(int &x : a) cin >> x;
    const int N = 1e5;
    vector<int> cntOdd(N+1), cntEven(N+1);
    bool isOdd = 1;
    for(int x : a){
        isOdd ? ++cntOdd[x] : ++cntEven[x];
        isOdd = !isOdd;
    }
    pair<int,int> maxOdd, maxEven;
    for(int i = 0; i < N; ++i){
        maxOdd = max(maxOdd, {cntOdd[i], i});
        maxEven = max(maxEven, {cntEven[i], i});
    }
    
    const int halfSize = n >> 1;    
    int ans1 = halfSize - maxOdd.first;
    pair<int,int> even;
    cntEven[maxOdd.second] = 0;
    for(int i = 1; i < n; i += 2){
        even = max(even, {cntEven[a[i]], a[i]});
    }
    ans1 += halfSize - even.first;

    int ans2 = halfSize - maxEven.first;
    pair<int,int> odd;
    cntOdd[maxEven.second] = 0;
    for(int i = 0; i < n; i += 2){
        odd = max(odd, {cntOdd[a[i]], a[i]});
    }
    ans2 += halfSize - odd.first;

    cout << min(ans1, ans2);
}