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.*;
public class Main {

  public static void main(String[] args) { 
      Scanner sc = new Scanner(System.in);
      int n = sc.nextInt();
      int k = sc.nextInt();
      String s = sc.next();
      List<Integer> arr = new ArrayList<>();
      for(int i = 0; i < s.length(); i++) {
        int flag = 1;
        if(s.charAt(i) == 'L')
          flag = -1;
        int cnt = 1;
        while(i < s.length() - 1 && s.charAt(i) == s.charAt(i+1)) {
          cnt++;
          i++;
        }
        arr.add(cnt * flag);
      }

      int[] leftSum = new int[arr.size()];
      for(int i = 1; i < arr.size(); i++)
        leftSum[i] = leftSum[i-1] + Math.abs(arr.get(i-1)) - 1;
        
      int[] rightSum = new int[arr.size()];
      for(int i = rightSum.length - 2; i >= 0; i--)
        rightSum[i] = rightSum[i+1] + Math.abs(arr.get(i+1)) - 1;

      int maxSum = 0;
      int replaceCnt = 0;
      int sum = 0;
      for(int i = 0, leftIdx = 0; i < arr.size(); i++) {
        int num = arr.get(i);
        if(num < 0) replaceCnt++;
        sum += Math.abs(num);
        while(replaceCnt > k) {
          int leftNum = arr.get(leftIdx++);
          if(leftNum < 0) replaceCnt--;
          sum -= Math.abs(leftNum);
        }
        
        maxSum = Math.max(maxSum, sum + leftSum[leftIdx] + rightSum[i] - 1);
      }

      replaceCnt = 0;
      sum = 0;
      for(int i = 0, leftIdx = 0; i < arr.size(); i++) {
        int num = arr.get(i);
        if(num > 0) replaceCnt++;
        sum += Math.abs(num);
        while(replaceCnt > k) {
          int leftNum = arr.get(leftIdx++);
          if(leftNum > 0) replaceCnt--;
          sum -= Math.abs(leftNum);
        }
        
        maxSum = Math.max(maxSum, sum + leftSum[leftIdx] + rightSum[i] - 1);
      }
      
      System.out.println(maxSum);
  }
}