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<stdlib.h>
char reverse(char s){
    if(s=='L')return 'R';
    else return 'L';
}
int main(){
    int n, k;
    char s[100000+10];
    char _s[100000+10];
    int score=0;
    int max=0;
    scanf("%d %d", &n, &k);
    scanf("%s", s);

    for(int i=0; i<n; i++){
        _s[i]=s[i];
    }
    for(int l=0; l<n-1; l++){
        for(int r=l+1; r<n; r++){
            for(int i=0; i<(r+1-l+1)/2; i++){
                char temp = s[l+i];
                s[l+i]=reverse(s[r-i]);
                s[r-i]=reverse(temp);
            }
            score=0;
            for(int i=0; i<n; i++){
                if(s[i]=='R' && i!=n-1){
                    if(s[i+1]=='R')score++;
                }else if(s[i]=='L' && i!=0){
                    if(s[i-1]=='L')score++;
                }
            }
            // printf("%d %d %s: %d\n", l, r, s, score);
            if(max<score)max=score;
        }
        for(int a=0; a<n; a++){
            s[a]=_s[a];
        }
    }

    printf("%d\n", max);
    return 0;

}