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)

'use strict';
const main = input => {
  const n = parseInt(input[0]), v = input[1].split(' ').map(val => parseInt(val));
  // console.log(n, v);
  let even = [], odd = [];
  for(let i=0; i<n; i++) {
    if(i%2) odd.push(v[i]);
    else even.push(v[i]);
  }
  // console.log(even, odd);

  let oddmode = (new Array(100000)).fill(0);
  odd.map(val => oddmode[val]++);
  let Omax = Math.max.apply(null, oddmode);
  let OmaxId = oddmode.indexOf(Math.max.apply(null, oddmode));
  oddmode[OmaxId] = 0;
  let Osec = Math.max.apply(null, oddmode);
  let OsecId = oddmode.indexOf(Math.max.apply(null, oddmode));
  
  let evenmode = (new Array(100000)).fill(0);
  even.map(val => evenmode[val]++);
  let Emax = Math.max.apply(null, evenmode);
  let EmaxId = evenmode.indexOf(Math.max.apply(null, evenmode));
  evenmode[EmaxId] = 0;
  let Esec = Math.max.apply(null, evenmode);
  let EsecId = evenmode.indexOf(Math.max.apply(null, evenmode));
  
  // console.log(oddmode, evenmode);
  // console.log(Omax, Emax, Osec, Esec);
  
  if (EmaxId != OmaxId) {
    console.log(n-Emax-Omax);
    return;
  }
  if (!Esec&&!Osec) {
    console.log(n/2);
    return;
  }
  if (!Esec) {
    console.log(n-Emax-Osec);
    return;
  }
  if (!Osec) {
    console.log(n-Esec-Omax);
    return;
  }
  console.log(Math.min(n-Emax-Osec, n-Esec-Omax));
  

  // if (OmaxId == EmaxId) {
  //   if (odd.length-Omax == even.length-Emax) {
  //     if (odd.length-Osec > even.length-Esec) {
  //       console.log(n-Esec-Omax);
  //     } else {
  //       console.log(n-Osec-Emax);
  //     }
  //   } else if (odd.length-Omax > even.length-Emax) {
  //     console.log(n-Omax-Esec);
  //   } else {
  //     console.log(n-Emax-Osec);
      
  //   }
  // } else {
  //   console.log(n-Omax-Emax);
  // }
  // // console.log(even.length, odd.length);
  
}
main(require("fs").readFileSync("/dev/stdin", "utf8").split('\n'));