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 Python (3.8.2) (judge verdict: Accepted)

N=int(input())
*V,=map(int,input().split())
import collections

import collections

c1=collections.Counter(V[0::2])
c1=sorted(c1.items(), key=lambda x: x[1], reverse=True)
e=c1[0][0]

c2=collections.Counter(V[1::2])
c2=sorted(c2.items(), key=lambda x: x[1], reverse=True)
o=c2[0][0]

cnt=0
if e!=o:
    cnt=sum(e!=i for i in V[0::2])
    cnt+=sum(o!=i for i in V[1::2])
else:
    if len(c1)==1 and len(c2)==1:
        cnt=N//2
    elif len(c1)==1:
        cnt1=sum(e!=i for i in V[0::2])
        cnt1+=sum(c2[1][0]!=i for i in V[1::2])
        
        cnt2=sum(e!=i for i in V[0::2])
        cnt2+=sum(c2[1][1]!=i for i in V[1::2])
        cnt=min(cnt1,cnt2)
    elif len(c2)==1:
        cnt1=sum(c1[1][0]!=i for i in V[0::2])
        cnt1+=sum(o!=i for i in V[1::2])

        cnt2=sum(c1[1][1]!=i for i in V[0::2])
        cnt2+=sum(o!=i for i in V[1::2])
        cnt=min(cnt1,cnt2)
    else:    
        cnt1=sum(e!=i for i in V[0::2])
        cnt1+=sum(c2[1][0]!=i for i in V[1::2])
        
        cnt2=sum(c1[1][0]!=i for i in V[0::2])
        cnt2+=sum(o!=i for i in V[1::2])
        cnt=min(cnt1,cnt2)
print(cnt)