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;
using ll = long;
main(){
  int n;cin>>n;
  vector<ll> a(n);
  for(auto&f:a)cin>>f;
  map<ll,ll> cnt1,cnt2;
  for(int i=0;i<n;i+=2)cnt1[a[i]]++;
  for(int i=1;i<n;i+=2)cnt2[a[i]]++;
  
  pair<ll,ll> cnt1_max(0,0),cnt1_next(0,0);
  pair<ll,ll> cnt2_max(0,0),cnt2_next(0,0);
  for(const auto&elem:cnt1){
    if(elem.second>cnt1_max.second){
      if(cnt1_max.second>cnt1_next.second)cnt1_next=cnt1_max;
      cnt1_max=elem;
    }
    else if(elem.second>cnt1_next.second)cnt1_next=elem;
  }
  for(const auto&elem:cnt2){
    if(elem.second>cnt2_max.second){
      if(cnt2_max.second>cnt2_next.second)cnt2_next=cnt2_max;
      cnt2_max=elem;
    }
    else if(elem.second>cnt2_next.second)cnt2_next=elem;
  }
  
  ll num=n/2;
  if(cnt1_max.first==cnt2_max.first){
    ll ans1=0,ans2=0;
    ans1+=num-cnt1_max.second;
    ans1+=num-cnt2_next.second;
    ans2+=num-cnt1_next.second;
    ans2+=num-cnt2_max.second;
    cout<<min(ans1,ans2)<<endl;
  }
  else {
    ll ans=0;
    ans+=num-cnt1_max.second;
    ans+=num-cnt2_max.second;
    cout<<ans<<endl;
  }
}