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 Kotlin (1.0.0) (judge verdict: Wrong Answer)

fun main(arr:Array<String>) {
    val itemCount = readLine()!!.toInt()
    val numbers = readLine()!!.split(" ").map { it.toInt() }
    val odd = mutableMapOf<Int,Int>()
    val even = mutableMapOf<Int,Int>()
    numbers.indices.forEach {
        if(it%2==0) {
            even[numbers[it]] = (even[numbers[it]]?:0) + 1
        } else {
            odd[numbers[it]] = (odd[numbers[it]]?:0) + 1
        }
    }
    var ans = itemCount
    val oddList = odd.toList().sortedByDescending { it.second }
    val evenList = even.toList().sortedByDescending { it.second }
    if(oddList.first().first == evenList.first().first) {
        ans = itemCount - oddList.first().second - if(evenList.size <= 1) 0 else evenList[1].second
        ans = Math.min(ans,itemCount - evenList.first().second - if(oddList.size <= 1) 0 else 1)
    } else {
        ans = itemCount - oddList.first().second - evenList.first().second
    }
}