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 Rust (1.15.1) (judge verdict: Accepted)

use std::vec::Vec;
use std::cmp::*;

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>().split_whitespace().map(|e| e.parse().ok().unwrap()).collect()
}

fn main() {
  solve_1()
}

fn solve_1()  {
  let (n, k) = {
    let v: Vec<usize> = read_vec();
    (v[0], v[1])
  };
  let mut s: Vec<char> = read::<String>().chars().collect();

  let mut score = 0;
  for i in 0..n-1 {
    if s[i] == s[i+1] {
      score += 1;
    }
  }

  let ans = min(score + 2*k, n-1);
  println!("{}", ans);
}

fn solve() {
  let (n, k) = {
    let v: Vec<usize> = read_vec();
    (v[0], v[1])
  };
  let mut s: Vec<char> = read::<String>().chars().collect();
  let mut k = k;
  let count = count_same_element(&s);
  for (i, &(left, right)) in count.iter().enumerate() {
    if i % 2 == 1 && k > 0 {
      let c = change(s[left]);
      k -= 1;
      for j in left..right+1 {
        s[j] = c;
      }
    }
  }
  let mut ans = 0;
  for i in 0..n-1 {
    if s[i] == s[i+1] {
      ans += 1;
    }
  }
  println!("{}", ans);
}

fn count_same_element<T: Eq + Copy>(v: &Vec<T>) -> Vec<(usize, usize)> {

  let mut res: Vec<(usize, usize)> = Vec::new();
  if v.len() == 0 {
    return res;
  }

  let mut now: T = v[0];
  let mut count: usize = 0;
  let mut left = 0;
  for (i, &e) in v.iter().enumerate() {
    if e == now {
      count += 1;
    }
    else {
      res.push((left, count+left-1));
      count = 1;
      now = e;
      left = i;
    }
  }
  res.push((left, count+left-1));

  res
}

fn change(c: char) -> char {
  match c {
    'R' => 'L',
    _ => 'R',
  }
}