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.util.Scanner;

public class Main {
    public static void main(String[] args) {
        try (Scanner sc = new Scanner(System.in)) {
            int N = sc.nextInt();
            int K = sc.nextInt();
            String S = sc.next();
            int[] s = new int[N];
            for (int i = 0; i < N; i++) {
                s[i] = S.charAt(i) == 'L' ? 0 : 1;
            }

            int score = score(s);
//            Utils.debug(score(s));
//            for (int l = 0; l < s.length; l++) {
//                for (int r = l + 1; r < s.length; r++) {
//                    reverse(s, l, r);
//                    Utils.debug(l, r, score(s));
//                    reverse(s, l, r);
//                }
//            }

            System.out.println(Math.min(N - 1, score + 2 * K));
        }
    }

    private static void reverse(int[] s, int l, int r) {
        for (int l2 = l, r2 = r; l2 < r2; l2++, r2--) {
            int swap = s[l2];
            s[l2] = s[r2];
            s[r2] = swap;
        }
        for (int i = l; i <= r; i++) {
            s[i] = 1 - s[i];
        }
    }

    private static int score(int[] s) {
        int score = 0;
        for (int i = 0; i < s.length; i++) {
            if (s[i] == 0 && (i - 1 >= 0 && s[i - 1] == 0)) {
                score++;
            }
            if (s[i] == 1 && (i + 1 < s.length && s[i + 1] == 1)) {
                score++;
            }
        }
        return score;
    }
}