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)

/// ITNOG :)))
/// SAliB
#include <bits/stdc++.h>

using namespace std;

#define pb                      push_back
#define pp                      pop_back
#define mp                      make_pair
#define fi                      first
#define sec                     second
#define xmax(x,y)               ((x) = max((x),(y)))
#define xmin(x,y)               ((x) = min((x),(y)))
#define all(x)                  (x).begin(),(x).end()
#define mem(x , y)              memset(x , y , sizeof(x))
#define _                       ios_base::sync_with_stdio(false),cin.tie(NULL),cout.tie(NULL);
#define debug1(a)               cout << string(40 , '*') << '\n' << #a << " ---> " << a << '\n';
#define debug2(a , b)           cout << string(40 , '*') << '\n' << #a << " ---> " << a << '\n' << #b << " ---> " << b << '\n';
#define debug3(a , b , c)       cout << string(40 , '*') << '\n' << #a << " ---> " << a << '\n' << #b << " ---> " << b << '\n' << #c << " ---> " << c << '\n';
#define debug4(a , b , c , d)   cout << string(40 , '*') << '\n' << #a << " ---> " << a << '\n' << #b << " ---> " << b << '\n' << #c << " ---> " << c << '\n' << #d << " ---> " << d << '\n';
#define connect(a , b , c)      a[b].pb(c) , a[c].pb(b)
//
typedef long long int ll            ;
typedef unsigned long long int ull  ;
typedef unsigned int uint           ;
typedef long double ld              ;
typedef string str                  ;
typedef pair < int , int > pii      ;
typedef pair < pii , int > ppi      ;
typedef pair < int , pii > pip      ;
typedef pair < ll , ll> pll         ;
typedef pair < ll , pll> plp        ;
typedef pair < pll , ll> ppl        ;
typedef vector < ll > vl            ;
typedef vector < ld > vld           ;
typedef vector < int > vi           ;
typedef vector < str > vs           ;
//
inline int   iin()  { int x   ; scanf("%d" , &x)   ;  return x; }
inline ll    lin()  { ll  x   ; scanf("%lld" , &x) ;  return x; }
inline ll    Lin()  { ll  x   ; scanf("%I64d" , &x);  return x; }
//
inline int   iout(int  x)  { printf("%d" , x)   ;  }
inline ll    lout(ll   x)  { printf("%lld" , x) ;  }
inline ll    Lout(ll   x)  { printf("%I64d" , x);  }
//
const int maxn4 = 1e4 + 85 , maxn5 = 1e5 + 69 , maxn6 = 1e6 + 85 , maxn7 = 1e7 + 69 , maxn8 = 1e8 + 85 , maxn9 = 1e9 + 69 , modi1 = 1e7 + 7 , modi2 = 1e9 + 7;
const ll mod1 = 1e7 + 7 , mod2 = 1e9 + 7;
int mn = INT_MAX , mx = INT_MIN;
//
int h[maxn5];
int main()
{_
    int n; cin >> n;
	int a[n];
	for(int i = 0 ; i < n ; i++)
	    cin >> a[i];
	int t1 = n , t3 = n , t2 = n , t4=n , t5 , t6;
	for(int i = 1 ; i < n ; i += 2)
		h[a[i]]++;
	for(int i = 0 ; i < maxn5 ; i++)
	{
		xmin(t1 , (n / 2) - h[i]);
		if(t3 > t1)
		{
			swap(t3 , t1);
			t5 = i;
		}
	}
	mem(h , 0);
	for(int i = 0 ; i < n ; i += 2)
		h[a[i]]++;
	for(int i = 0 ; i < maxn5 ; i++)
	{
		xmin(t2 , (n / 2) - h[i]);
		if(t4 > t2)
		{
			swap(t4 , t2);
			t6 = i;
		}
	}
	if (t5 != t6) 
	    printf("%d" , t3 + t4);
	else
		printf("%d",min(t1 + t4 , t3 + t2));
	return 0;
}