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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int main(void) {
	int ve[50000];
	int vo[50000];
	int n,i,j,c,checke,checko,l;
	int counte[100000];
	int counto[100000];
	scanf("%d",&n);
	for(i=0;i<n;i++){
		if((i%2)==0){
			scanf("%d",&ve[i/2]);
		}else{
			scanf("%d",&vo[(i-1)/2]);
		}
	}
	for(i=0;i<100000;i++){
		counte[i]=0;
		counto[i]=0;
	}
	for(i=0;i<((n/2)+1);i++){
		for(j=0;j<((n/2)+1);j++){
			if(i>j){
				if(ve[i]==ve[j]){
					counte[ve[i]]++;
				}
				if(vo[i]==vo[j]){
					counto[vo[i]]++;
				}
				if((ve[i]==vo[j])||(ve[j]==vo[i])) c++;
			}
		}
	}
	checke=0;
	checko=0;
	for(i=0;i<99999;i++){
		if(counte[i]>checke){
			if(checke==0){
				checke=counte[i];
			}else{
				checke=counte[i];
				c++;
			}
		}
		if(counto[i]<counto[i+1]){
			if(checko==0){
				checko=counto[i];
			}else{
				checko=counto[i];
				c++;
			}
		}
	}
	l=checke+checko;
	for(i=0;i<100000;i++){
		checke=checke-counte[i];
		checko=checko-counto[i];
	}
	if(checke<0) checke=-checke;
	if(checko<0) checko=-checko;
	c=c+checke+checko-l;
	printf("%d",c);
	return 0;
}