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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
int main(void)
{
	int n,v[100001],a[100001],b[100001],i,x1=0,x2=0,y1,y2,z1=0,z2=0,ans;
	scanf("%d",&n);
	for(i=0; i<n; i++){
		scanf("%d",&v[i]);
	}
	for(i=0; i<n; i+=2){
		if(a[v[i]]>0){
			a[v[i]]++;
		}else{
			a[v[i]]=1;
		}
		if(x1<a[v[i]]){x1=a[v[i]];y1=v[i];}
	}
	for(i=1; i<n; i+=2){
		if(b[v[i]]>0){
			b[v[i]]++;
		}else{
			b[v[i]]=1;
		}
		if(x2<b[v[i]]){x2=b[v[i]];y2=v[i];}
	}
	if(y1!=y2){
		ans=(n/2-x1)+(n/2-x2);
	}else{
		if(x1<x2){
			for(i=0; i<n; i+=2){
				if(z1<a[v[i]] && v[i]!=y1)z1=a[v[i]];
			}
			ans=(n/2-z1)+(n/2-x2);
		}else if(x1>x2){
			for(i=1; i<n; i+=2){
				if(z2<b[v[i]] && v[i]!=y2)z2=b[v[i]];
			}
			ans=(n/2-x1)+(n/2-z2);
		}else{
			for(i=0; i<n; i+=2){
				if(z1<a[v[i]] && v[i]!=y1)z1=a[v[i]];
			}
			for(i=1; i<n; i+=2){
				if(z2<b[v[i]] && v[i]!=y2)z2=b[v[i]];
			}
			if(z1>=z2){
				ans=(n/2-z1)+(n/2-x2);
			}else{
				ans=(n/2-x1)+(n/2-z2);
			}
		}
	}
	printf("%d\n",ans);
	return 0;
}