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 Kotlin (1.3.71) (judge verdict: Wrong Answer)

import java.io.PrintWriter

fun main() {
    PrintWriter(System.out).use { out ->
        solve(out)
        out.flush()
    }
}

fun solve(out: PrintWriter) {
    var (_, k) = readIntegerList()
    val s = readLine()!!
    val runLength = runLengthEncode(s)
    var unHappy = 0
    for (i in 1..runLength.lastIndex) {
        if (runLength[i].value == 'L' && runLength[i - 1].value == 'R') {
            unHappy += 2
        }
    }

    var edge = 0
    if (runLength.first().value == 'L') {
        edge++
    }

    if (runLength.last().value == 'R') {
        edge++
    }

    while (k > 0 && unHappy > 0) {
        unHappy -= 2
        k--
    }

    if (k >= 2) {
        edge = 0
    } else if (k == 1 && edge >= 1) {
        edge--
    }

    println(s.length - unHappy - edge)
}


data class RunLength<T>(val value: T, val length: Int)

/** O(s.length) */
fun runLengthEncode(s: CharSequence): List<RunLength<Char>> {
    if (s.isEmpty()) return emptyList()

    val runLengthList = mutableListOf<RunLength<Char>>()
    var currentChar = s.first()
    var length = 1
    for (c in s.substring(1)) {
        if (c == currentChar) {
            length++
        } else {
            runLengthList += RunLength(currentChar, length)
            currentChar = c
            length = 1
        }
    }
    runLengthList += RunLength(currentChar, length)

    return runLengthList
}

fun readInteger() = readLine()!!.toInt()
fun readLong() = readLine()!!.toLong()
fun readStringList() = readLine()!!.split(" ")
fun readIntegerList() = readStringList().map(String::toInt)
fun readLongList() = readStringList().map(String::toLong)
fun readDoubleList() = readStringList().map(String::toDouble)