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 rep(i,n) for(int i = 0; i < (n); i++)
using namespace std;
using ll = long long;
using P = pair<int, int>;

int main() {
  int n;
  cin >> n;
  vector<int> v(n);
  rep(i,n) cin >> v.at(i);
  map<int, int> mp1;
  for(int i = 0; i <= n-1; i += 2) {
    mp1[v.at(i)]++;
  }
  int a;
  int max1 = 0;
  for(auto p : mp1) {
    max1 = max(max1, p.second);
    if(max1==p.second) a = p.first;
  }
  map<int, int> mp2;
  for(int i = 1; i <= n-1; i += 2) {
    mp2[v.at(i)]++;
  }
  int max2 = 0;
  for(auto p : mp2) {
    if(p.first==a) continue;
    max2 = max(max2, p.second);
  }
  int ans1 = n - max1 - max2;
  int b;
  int max3 = 0;
  for(auto p : mp2) {
    max3 = max(max3, p.second);
    if(max3==p.second) b = p.first;
  }
  int max4 = 0;
  for(auto p : mp1) {
    if(p.first==b) continue;
    max4 = max(max4, p.second);
  }
  int ans2 = n - max3 - max4;
  cout << max1 << endl;
  cout << max2 << endl;
  cout << max3 << endl;
  cout << max4 << endl;
  cout << min(ans1,ans2) << endl;
  return 0;
}