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: Runtime Error)

#include<stdio.h>

int main(){
		int n,i,j,num[2][50001][2]={-1},a,counter[2]={0},flag=0;
		scanf("%d",&n);
		for(i=0;i<n;i++){
				scanf("%d",&a);
				flag=0;
				for(j=0;j<counter[i%2];j++){
						if(num[i%2][j][0]==a){
								num[i%2][j][1]++;
								flag=1;
						}
				}
				if(flag==0){
						num[i%2][counter[i%2]][0]=a;
						num[i%2][counter[i%2]][1]=1;
						counter[i%2]++;
				}
		}
		int maxO[2][2]={0},maxE[2][2]={0};
		for(i=0;i<n;i++){
				if(num[0][i][1]>0){
				if(maxO[0][1]<num[0][i][1]){
						maxO[0][0]=num[0][i][0];
						maxO[0][1]=num[0][i][1];
				}
				else{
						if(maxO[1][1]<num[0][i][1]){
								maxO[1][0]=num[0][i][0];
								maxO[1][1]=num[0][i][1];	
						}
				}
				}
				if(num[1][i][1]>0){
				if(maxE[0][1]<num[1][i][1]){
						maxE[0][0]=num[1][i][0];
						maxE[0][1]=num[1][i][1];
				}
				else{
						if(maxO[1][1]<num[1][i][1]){
								maxE[1][0]=num[1][i][0];
								maxE[1][1]=num[1][i][1];	
						}
				}
				}
		}
		if(maxO[0][0]!=maxE[0][0]){
	//			puts("OK");
	//			printf("%d %d\n",maxO[0][1],maxE[0][1]);
				printf("%d",n-maxE[0][1]-maxO[0][1]);
		}
		else {
//				puts("OK1");
//				printf("%d %d\n",num[0][0][0],num[0][1][0]);
				int ans=n-maxO[1][1]-maxE[0][1];
				if(ans>n-maxO[0][1]-maxE[1][1])ans=n-maxO[0][1]-maxE[1][1];
				printf("%d",ans);
		}
		return 0;
}