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# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;

namespace AtCoder
{
	public class Program
	{
		// ABC111-C
		static void Main(string[] args)
		{
			Dictionary<string, int> odd = new Dictionary<string, int>();
			Dictionary<string, int> even = new Dictionary<string, int>();

			int n = int.Parse(Console.ReadLine());
			string[] vs = Console.ReadLine().Split((string[])null, StringSplitOptions.RemoveEmptyEntries);
			for (int i = 0; i < n; i++) {
				if (i % 2 != 0) {
					if (!odd.ContainsKey(vs[i])) {
						odd[vs[i]] = 0;
					}
					odd[vs[i]]++;
				} else {
					if (!even.ContainsKey(vs[i])) {
						even[vs[i]] = 0;
					}
					even[vs[i]]++;
				}
			}

			string odd1Str = "", odd2Str = "", even1Str = "", even2Str = "";
			int odd1Cnt = 0, odd2Cnt = 0, even1Cnt = 0, even2Cnt = 0;
			FindMax(odd, ref odd1Str, ref odd1Cnt, ref odd2Str, ref odd2Cnt);
			FindMax(even, ref even1Str, ref even1Cnt, ref even2Str, ref even2Cnt);

			int result = 0;
			if (odd1Str != even1Str) {
				// 奇数の最頻値と偶数の最頻値が違う
				result = n - odd1Cnt - even1Cnt;
			} else {
				// 奇数の最頻値と偶数の最頻値が同じ
				if (odd1Cnt + even2Cnt > even1Cnt + odd2Cnt) {
					result = n - odd1Cnt - even2Cnt;
				} else {
					result = n - even1Cnt - odd2Cnt;
				}
			}
			Console.WriteLine(result);
		}

		private static void FindMax(Dictionary<string, int> dic, ref string max1Str, ref int max1Cnt, ref string max2Str, ref int max2Cnt)
		{
			foreach (var item in dic) {
				if (item.Value > max1Cnt) {
					max2Str = max1Str;
					max2Cnt = max1Cnt;
					max1Str = item.Key;
					max1Cnt = item.Value;
				} else if (item.Value > max2Cnt) {
					max2Str = item.Key;
					max2Cnt = item.Value;
				}
			}
		}
	}
}