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: Accepted)

import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.*;
import java.lang.Math.*;
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        HashMap<Integer, Integer> mem = new HashMap<>();
        LinkedList<Integer> dic = new LinkedList<>();
        int n = in.nextInt();
        int k = in.nextInt();
        String s = in.next();
        char p = s.charAt(0);
        int c=1;
        for(int i=1;i<n;i++){
            if(s.charAt(i)==p){
                c++;
            }
            else {
                dic.add(c);
                c=1;
                p=s.charAt(i);
            }

        }
        dic.add(c);
//        System.out.println(dic);
        int count =0;
        int sum =0;
        while (k>0 && dic.size()>0){
            sum+=dic.removeFirst();
            count++;
            if (count==3){
                k--;
                count=1;
            }
        }
        int ans=0;
        ans+= sum-1;
        while (dic.size()>0){
            ans+=dic.removeFirst()-1;
        }
        System.out.println(ans);
    }
}