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.math.BigInteger;
import java.util.Scanner;
import java.util.Vector;

public class Main {
	public static void main(String[] args) {
		@SuppressWarnings("resource")
		Scanner scanner = new Scanner(System.in);

		int N = scanner.nextInt();

		int K = scanner.nextInt();

		String S = scanner.next();

		String[] scannerArray = new String[N];

		scannerArray = S.split("");

		long ans = 0;

		for(int index = 1; index < N; index++) {
			if(scannerArray[index - 1].equals(scannerArray[index])) {
				ans++;
			}
		}

		ans += 2*K;

		log(Math.min(ans, N-1));

	}

	private static void dfs(int[] treeArray,int parrent, int children) {
		Vector<Vector> aa = new Vector<>();
		Vector<Integer> bb = new Vector<>();
		bb.add(1);
		aa.add(bb);
	}


	/**
	 * バブルソートの交換数を算出
	 * ※転倒数
	 * @param array
	 * @return
	 */
	private static long countSwapBabbleSort(int[] array, int N) {
		long ans = 0;
		int[] binary = new int[N + 1];

		for (int index = 0; index < array.length; index++) {
			ans = ans + (index - sum(binary, array[index]));
			add(binary, array[index], 1);
		}
		return ans;
	}

    //add value at idx on bit O(logN)
	private static void add(int[] bit,int idx,int value) {
        for(int i=idx;i<bit.length;i=i+(i&-i)) {
        	bit[i] += value;
        }
    }

    //return sum [1,idx] O(logN)
	private static long sum(int[] bit,int idx) {
        long ret = 0;
        for(int i=idx;i>0;i-=(i&-i)) {
        	ret += bit[i];
        }
        return ret;
    }

	private static void log(String str) {
		System.out.println(str);
	}

	private static void log(int str) {
		System.out.println(str);
	}

	private static void log(long str) {
		System.out.println(str);
	}

	private static void log(double str) {
		System.out.println(str);
	}

	private static void log(BigInteger str) {
		System.out.println(str);
	}

}