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<iostream>
#include<map>
#include<algorithm>

using namespace std;
int main(){
    int n;
    cin >> n;
    int v[100000];
    for (int i = 0; i < n; i++){
        cin >> v[i];
    }
    map<int, int> odd;
    map<int, int> even;
    for (int i = 0; i < n; i = i + 2){
        even[v[i]]++;
    }
    for (int i = 1; i < n; i = i + 2){
        odd[v[i]]++;
    }
    int maxeven = 0;
    int maxevenkey;
    int secondmaxeven = 0;
    for (auto itr = even.begin(); itr != even.end(); itr++){
        if (maxeven < itr->second) {
            maxeven = itr->second;
            maxevenkey = itr->first;
        }
        else if (itr->second < maxeven && secondmaxeven < itr->second) secondmaxeven = itr->second;
    }

    int maxodd = 0;
    int maxoddkey;
    int secondmaxodd = 0;    
    for (auto itr = odd.begin(); itr != odd.end(); itr++){
        if (maxodd < itr->second) {
            maxodd = itr->second;
            maxoddkey = itr->first;
        }
        else if (itr->second < maxodd && secondmaxodd < itr->second) secondmaxodd = itr->second;
    }

    if (maxevenkey != maxoddkey){
        cout << n - maxodd - maxeven << endl;
    }
    else{
        cout << min(n - secondmaxodd - maxeven, n - maxodd - secondmaxeven) << endl;
    }
    
}