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>

int main(int argc, const char * argv[]) {
    int N, K;
    int pos,flag,pos_a,cnt;
    pos=flag=pos_a=cnt=0;
    char now;
    scanf("%d%d", &N, &K);
    scanf("%c", &now);
    now = ' ';
    char S[N];
    for(int i=0;i<N;i++){
        scanf("%c", &S[i]);
    }
    for(int i=1;i<N-1;i++){
        if((S[i-1] != S[i]) && flag == 0){
            now=S[i];
            pos = i;
            flag =1;
        }else if((S[i-1] != S[i]) && flag == 1){
            pos_a=i-1;
            for(int j=0;j<(pos_a-pos+1);j++){
                if(S[pos+j] == 'R'){
                    S[pos+j] = 'L';
                    now = ' ';
                }else{
                    S[pos+j] = 'R';
                    now = ' ';
                }
            }
            flag=0;
            cnt++;
        }
        if(cnt >= K)break;
    }
    if(cnt < K){
        if(now != ' '){
            for(int i=pos;i<N;i++){
                if(now == 'R'){
                    S[i] = 'L';
                }else{
                    S[i] = 'R';
                }
            }
        }
        else if(S[N-1] == 'R'){
            S[N-1] = 'L';
        }else{
            S[N-1] = 'R';
        }
    }
    int ans = 0;
    for(int i=0;i<N-1;i++){
        if(S[i] == S[i+1]){
            ans++;
        }else{
            continue;
        }
    }
    printf("%d\n", ans);
    return 0;
}