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.*;
public class Main{
    public static void main(String[] args){
      Scanner sc = new Scanner(System.in);
      int n = sc.nextInt();
      int k = sc.nextInt();
      char[] c = sc.next().toCharArray();
      int cntR = 0;
      int cntL = 0;
      int res = 0;
      char x = 'a';
      char y = 'a';
      for(int i = 0;i < n;i++){
      	if(c[i]=='R')cntR++;
        else cntL++;
      }
      if(cntR>=cntL){
        x = 'R';
        y = 'L';
      }else{
      	x = 'L';
        y = 'R';
      }
      for(int i = 0;i < n-1;i++){
      	if(c[i]=='R'&&c[i]==c[i+1])res++;
      }
      for(int i = 1;i < n;i++){
      	if(c[i]=='L'&&c[i]==c[i-1])res++;
      }
      
      for(int i = 0;i < c.length-1&&k > 0;i++){
        if(c[i]==x&&c[i+1]==y){
            res+=2;
            k--;
        }
        if(i==0&&x=='L')res--;
        if(i==n-1&&x=='R')res--;
      }
      
      if(c[0]=='L'&&x=='R'&&k > 0){
        k--;
        res++;
      }
      if(c[n-1]=='L'&&x=='L'&&k > 0){
        k--;
        res++;
      }
      System.out.println(res);
    }
}