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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;

class Main {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int n = Integer.parseInt(sc.next());
    int k = Integer.parseInt(sc.next());
    String s = sc.next();
    String c = "";
    int d[] = new int[n];
    int cnt = 0;
    int same = 0;
    int ans = 0;
    int ss = 0;
    
    for (int i=0; i<n-1; i++) {
      c = s.substring(i,i+1);
      d[i] = (c.equals("L")) ? -1 : 1;
    }
    
    for (int i=1; i<n; i++) {
      if(d[i] == 1){
        if(ss == 0) {
          ss++;
          same++;
          continue;
        }
        if(d[i] == d[i-1]) {
          same++;
        } else {
          for (int j=0; j<=same; j++){
            d[i-j] *= -1;
          }
          ss = 0;
          same = 0;
          cnt++;
          if(cnt == k) {
            same = 0; 
            break;
          }
        }
      }
      
    }

    for (int i=1; i<n; i++) {
      if(d[i] == d[i-1]){
        same++; 
      } else {
        ans += same;
        same = 0;
      }
      //System.out.print(d[i] == -1 ? "L" : "R");
    }
    
    System.out.println(ans);
  }
}