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: Wrong Answer)

use std::io::*;
use std::str::FromStr;

pub fn read<T: FromStr>() -> T {
  let stdin = stdin();
  let stdin = stdin.lock();
  let token: String = stdin
    .bytes()
    .map(|c| c.expect("failed to read char") as char)
    .skip_while(|c| c.is_whitespace())
    .take_while(|c| !c.is_whitespace())
    .collect();
  token.parse().ok().expect("failed to parse token")
}

pub fn read_n<T: FromStr>(n: usize) -> Vec<T> {
  read_n_logic::<T>(n, vec![])
}

pub fn read_n_logic<T: FromStr>(n: usize, mut a: Vec<T>) -> Vec<T> {
  match n {
    0 => a,
    _ => {
      a.push(read::<T>());
      read_n_logic(n - 1, a)
    }
  }
}

use std::cmp::{max, min};

fn main() {
  let n: usize = read();
  let k: usize = read();
  let s: String = read();
  let s_vec: Vec<char> =  s.chars().collect();

  let mut s_sum = vec![];
	let mut ans = 0;

  let mut cnt = 0;
  let mut cc = 'L';
  for (idx, i) in s_vec.iter().enumerate() {
      if cnt == 0 {
          cnt += 1;
          cc = *i;
          continue
      }
      if cc == *i {
          cnt += 1;
      } else {
          s_sum.push(cnt);
          cnt = 1;
          cc = *i;
      }

      if idx == s_vec.len() -1 {
          s_sum.push(cnt);
      }
  }

	// println!("s_sum: {:?}", s_sum);
  let bl = ((k + 1) * 2 -1) as i64 ;
	// println!("bl: {}", bl);
	if s_sum.len() as i64 - bl <= 0 {
		// println!("sum return");
		println!("{}", s_sum.iter().fold(0, |sum, i| sum + i) -1);
		return
	}

	let mut prev = 0;
	let mut remain = 0;
	for ki in 0..bl {
		prev += s_sum[ki as usize];
	}

	for ki in bl..s_sum.len() as i64 {
		remain += s_sum[ki as usize] - 1;
	}

	ans = prev -1 + remain;
	// println!("ans: {}", ans);
	// println!("remain: {}, prev: {}", remain, prev);

	for i in 0..s_sum.len() as i64 - bl {
		let next = i + bl;
		let front = s_sum[next as usize];
		let back = s_sum[i as usize];
		prev -= back;
		prev += front;

		remain += back -1;
		remain -= front -1;

		ans = max(ans, prev -1 + remain);
		// println!("remain: {}, prev: {}", remain, prev);
		// println!("ans: {}", ans);
	}

	println!("{}", ans);
}