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 Go (1.6) (judge verdict: Runtime Error)

package main

import (
	"bufio"
	"fmt"
	"os"
	"strconv"
	"strings"
)

var sc = bufio.NewScanner(os.Stdin)

func newline() string {
	sc.Scan()
	return sc.Text()
}

func main()  {
	in := newline()
	n, _ := strconv.Atoi(in)

	valArray := newline()
	vals := strings.Split(valArray, " ")

	odd := make(map[int]int)
	even := make(map[int]int)
	maxOdd, maxOddVal, maxEven, maxEvenVal := 0, 0, 0, 0
	priorMaxOdd, priorMaxEven := 0, 0
	for i := 0; i < n; i++ {
		thisVal, _ := strconv.Atoi(vals[i])
		if i % 2 == 0 {
			odd[thisVal]++
			if odd[thisVal] >= maxOddVal {
				maxOddVal = odd[thisVal]
				if maxOdd != thisVal{
					priorMaxOdd = maxOdd
				}
				maxOdd = thisVal
			}
		} else {
			even[thisVal]++
			if even[thisVal] >= maxEvenVal {
				maxEvenVal = even[thisVal]
				if maxEven != thisVal{
					priorMaxEven = maxEven
				}
				maxEven = thisVal
			}
		}
	}
	if maxOdd == maxEven {
		if odd[priorMaxOdd] + maxEvenVal < even[priorMaxEven] + maxOddVal {
			fmt.Println(outVal(n, maxOddVal, even[priorMaxEven]))
		} else {
			fmt.Println(outVal(n, odd[priorMaxOdd], maxEvenVal))
		}
	}else {
		fmt.Println(outVal(n, maxOddVal, maxEvenVal))
	}
}

func outVal(n, odd, even int) int {
	return n - odd - even
}