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: Accepted)

"use strict";
    
const main = arg => {
    arg = arg.trim().split("\n");
    
    const N    = ~~arg[0].split(" ")[0];
    const v    = arg[1].split(" ").map(n=>~~n);
    
    let even = [];
    let odd  = [];
    
    for(let i=0; i<N; i++) {
        const h = v[i];
        let 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).reverse().filter(a=>a);
    odd  = odd.sort((a,b)=>a.count-b.count).reverse().filter(a=>a);
  
    if(even[0].val!=odd[0].val){
        console.log(N - even[0].count - odd[0].count);
        return;
    }
    
    // 全て同じ値のとき
    if(!even[1] && !even[2]){
        console.log(N/2);
        return;
    }
    
    // 偶数番目のみ全て同じ値のとき
    if(!even[1]){
        console.log(N - even[0].count - odd[1].count);
        return;
    }
    
    // 奇数番目のみ全て同じ値のとき
    if(!odd[1]){
        console.log(N - even[1].count - odd[0].count);
        return;
    }
    
    console.log(Math.min(
        N - even[0].count - odd[1].count,
        N - even[1].count - odd[0].count
    ));
}
main(require('fs').readFileSync('/dev/stdin', 'utf8'));