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(args: Array<String>) {
    val n = readLine()!!.toLong()
    val aList = readLine()!!.split(' ').map(String::toLong)
    val evenList = aList.filterIndexed { index, l -> index % 2 == 0 }
    val oddList = aList.filterIndexed { index, l -> index % 2 == 1 }
    val evenGroupBy = evenList.groupBy { it }.toSortedMap().toList().sortedByDescending { it.second.size }
    val oddGroupBy = oddList.groupBy { it }.toSortedMap().toList().sortedByDescending { it.second.size }

    var count1 = 0L
    count1 += evenList.size - evenGroupBy[0].second.size
    var flag1 = false
    for (i in oddGroupBy) {
        if (i.first != evenGroupBy[0].first) {
            count1 += oddList.size - i.second.size
            flag1 = true
        }
    }
    if (!flag1) count1 += oddList.size

    var count2 = 0L
    count2 += oddList.size - oddGroupBy[0].second.size
    var flag2 = false
    for (i in evenGroupBy) {
        if (i.first != oddGroupBy[0].first) {
            count2 += evenList.size - i.second.size
            flag2 = true
        }
    }
    if (!flag2) count2 += evenList.size

    println(Math.min(count1, count2))
}