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 (1.42.0) (judge verdict: Accepted)

use std::io::Read;

fn main() {
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();

    let answer = solve(&buf);

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

fn solve(input: &str) -> String {
    let mut iterator = input.split_whitespace();

    let n: usize = iterator.next().unwrap().parse().unwrap();
    let k: usize = iterator.next().unwrap().parse().unwrap();
    let mut s: Vec<char> = iterator.next().unwrap().chars().collect();

    let mut prev_end = 0;
    for _ in 0..k {
        let mut begin = 0;
        for i in (prev_end + 1)..n {
            if s[i - 1] != s[i] {
                begin = i;
                break;
            }
        }
        if begin == 0 {
            return (n - 1).to_string();
        }
        let mut end = begin;
        for i in (begin + 1)..n {
            if s[i - 1] == s[i] {
                end = i;
            } else {
                break;
            }
        }
        for i in begin..=end {
            if s[i] == 'L' {
                s[i] = 'R';
            } else {
                s[i] = 'L';
            }
        }
        prev_end = end;
    }

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