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 JavaScript (node.js v5.12) (judge verdict: Wrong Answer)

// 解説をそのまま写経した感じです

function Main(input){
  input = input.split("\n");
  var n = ~~input[0]
  var v = input[1].split(" ");
  var even = [];
  var odd = [];
  for(var i = 0; i < n; i++){
  	var h = v[i];
    var a = (i%2?odd:even)[h];
    a = a || {val:h,count:0};
    a.count++;
    if(i%2){odd[h]=a}else{even[h]=a};
  }
  even = even.sort((a,b)=>a.count-b.count).filter(a=>a);
  odd = odd.sort((a,b)=>a.count-b.count).filter(a=>a);
  // filter : nullを消去
  
  if(even[0].val!=odd[0].val){
  	return n - even[0].count - odd[0].count;
    // 最頻値が異なる
  }
  if(!even[1] && !even[2]){
  	return n/2;
    // 偶奇どちらの副最頻値も存在しない
    // （すべての数字が同じ）
  }else if(!even[1]){
  	return n - even[0].count - odd[1].count;
    // 偶数番目の副最頻値が存在しない
  }else if(!odd[1]){
  	return n - even[1].count - odd[0].count;
    // 奇数番目の副最頻値が存在しない
  }else{
    return Math.min(
    	 n - even[0].count - odd[1].count,
       n - even[1].count - odd[0].count
    );
  }
}

console.log(Main(require("fs").readFileSync("/dev/stdin", "utf8")));