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 <cstdio>
#include <algorithm>
#include <limits>
#include <numeric>
#include <type_traits>
#include <iostream>

typedef long long ll;
typedef long long int lli;

#define rep(i,n) for(int i = 0; i < (int)(n); i++)
#define repi(i,a,b) for(int i=int(a);i<int(b);++i)

template<class T>bool umax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool umin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

using namespace std;

int v[200000],a[200000]={0},b[200000]={0};
int main() {
    int N ,mem_a=0, mem_b=0;
    cin>> N;

    for(int i=0; i<N; i=i+2){
        cin >> v[i] >> v[i+1];
        a[v[i]]++;
        b[v[i+1]]++;
    }

    int Max_a = 0;
    for(int i=0; i<100005; i++){
        if(Max_a<a[i]) {
            Max_a = a[i];
            mem_a=i;
        }
    }

    int Max_b = 0;
    for(int i=0; i<100005; i++){
        if(Max_b<b[i]) {
            Max_b = b[i];
            mem_b=i;
        }
    }

    int Sec_a = 0;
    a[mem_a]=0;
    for(int i=0; i<100005; i++){
        if(Sec_a<a[i]) {
            Sec_a = a[i];
        }
    }

    int Sec_b = 0;
    b[mem_b]=0;
    for(int i=0; i<100005; i++){
        if(Sec_b<b[i]) {
            Sec_b = b[i];
        }
    }

    int ans=0;
    if(mem_a==mem_b) {
        ans = min(N-Sec_a-Max_b, N-Sec_b-Max_a);
    }else{
        ans = N - Max_a - Max_b;
    }

    printf("%d", ans);
    return 0;

}