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::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() {
    let (n, k) = {
      let v: Vec<usize> = read_vec();
      (v[0], v[1])
    };
    let mut cs: Vec<char> = read::<String>().chars().collect();
    let mut s = cs.clone();
    let mut left_vec = Vec::new();
    let mut left = 0;
    let mut right = 0;
    let mut c = 'R';
    s.push(c);
    for i in 0..n+1 {
      if s[i] == c {
        right = i;
      }

      if left < right {
        if i != n {
            left_vec.push((left, right, right - left - 1));
        } else if right - left - 1 > 2 {
            left_vec.push((left, right, right - left - 1));
        } else if left == 0 && s[0]  != c {
            left_vec.push((left, right, right - left));
        }
        left = right;
      }
    }
    //println!("{:?}", s);
    //println!("{:?}", left_vec);

    left_vec.sort_by(|a, b| b.2.cmp(&a.2));
    //println!("{:?}", left_vec);

    for i in 0..min(k, left_vec.len()) {
      let (left, right, _ ) = left_vec[i];
      for j in left+1..right {
        s[j] = c;
      }
    }

    //println!("{:?}", s);

    let mut ans_1: usize = 0;
    for i in 0..n-1 {
      if s[i] == s[i+1] {
        ans_1 += 1;
      }
    }

    let mut s = cs.clone();
    let mut left_vec = Vec::new();
    let mut left = 0;
    let mut right = 0;
    let mut c = 'L';
    s.push(c);
    for i in 0..n+1 {
      if s[i] == c {
        right = i;
      }

      if left < right {
        if i != n {
            left_vec.push((left, right, right - left - 1));
        } else if right - left - 1 > 2 {
            left_vec.push((left, right, right - left - 1));
        }  else if left == 0 && s[0]  != c {
            left_vec.push((left, right, right - left));
        }
        left = right;
      }
    }
    //println!("{:?}", s);
    //println!("{:?}", left_vec);

    left_vec.sort_by(|a, b| b.2.cmp(&a.2));
    //println!("{:?}", left_vec);

    for i in 0..min(k, left_vec.len()) {
      let (left, right, _ ) = left_vec[i];
      for j in left+1..right {
        s[j] = c;
      }
    }

    //println!("{:?}", s);

    let mut ans_2: usize = 0;
    for i in 0..n-1 {
      if s[i] == s[i+1] {
        ans_2 += 1;
      }
    }
    println!("{}", max(ans_1, ans_2));
}