Problem:
Score : 400 points

Problem Statement

There are N people standing in a queue from west to east.

Given is a string S of length N representing the directions of the people.
The i-th person from the west is facing west if the i-th character of S is L, and east if that character of S is R.

A person is happy if the person in front of him/her is facing the same direction.
If no person is standing in front of a person, however, he/she is not happy.

You can perform the following operation any number of times between 0 and K (inclusive):

Operation: Choose integers l and r such that 1 \leq l \leq r \leq N, and rotate by 180 degrees the part of the queue: the l-th, (l+1)-th, ..., r-th persons. That is, for each i = 0, 1, ..., r-l, the (l + i)-th person from the west will stand the (r - i)-th from the west after the operation, facing east if he/she is facing west now, and vice versa.

What is the maximum possible number of happy people you can have?

Constraints

N is an integer satisfying 1 \leq N \leq 10^5.

K is an integer satisfying 1 \leq K \leq 10^5.

|S| = N

Each character of S is L or R.

Input

Input is given from Standard Input in the following format:

N K
S

Output

Print the maximum possible number of happy people after at most K operations.

Sample Input 1

6 1
LRLRRL

Sample Output 1

3

If we choose (l, r) = (2, 5), we have LLLRLL, where the 2-nd, 3-rd, and 6-th persons from the west are happy.

Sample Input 2

13 3
LRRLRLRRLRLLR

Sample Output 2

9

Sample Input 3

10 1
LLLLLRRRRR

Sample Output 3

9

Sample Input 4

9 2
RRRLRLRLL

Sample Output 4

7

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <string.h>

int hanten(char S[100099],int a,int b,int j){
	int len = strlen(S);
	int i;
	for(i=a;i<=b;i++)
	{
		if(S[i] == 'R') S[i] = 'L';
		else S[i] = 'R';
	}
	char tmp;
	int c = (b-a+1)/2;
	if((b-a+1)%2 == 1) c++;
	for(i=0;i<c;i++){
		tmp = S[a+i];
		S[a+1] = S[b-i];
		S[b-1] = tmp;
	}
	j++;
	return j;
}
int main(void)
{
	int N,K;
	scanf("%d %d",&N,&K);
	char S[N];
	scanf("%s",S);
	int i;
	int ans = 0;
	for(i=0;i<N;i++){
		if(S[i] == 'L' && S[i-1] == 'L') ans++;
		else if(S[i] == 'R' && S[i+1] == 'R') ans++;
	}
	int j = 0;
	int h;
	while(j<K){
		for(i=0;i<N-1;i++){
			//+2 sagasu;
			for(h=i+1;h<N&&((S[i]=='L'&&S[i-1] != 'L')||(S[i]=='R'&&S[i+1] != 'R'));h++)
			{
				if(S[i]==S[h] &&((S[h]=='L'&&S[h-1] != 'L')||(S[h]=='R'&&S[h+1] != 'R')))
				{
					j = hanten(S,i,h,j);
					ans += 2;
				}
				if(j >= K) break;
			}
			if(j >= K) break;
		}
	}
	printf("%d\n",ans);
	return 0;
}