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 JavaScript (node.js v5.12) (judge verdict: Accepted)

function Main(input) {
  input = input.split("\n");
  const n = input[0].split(" ")[0];
  const k = input[0].split(" ")[1];
  const s = input[1].split("");

  var happyRCount = 0;
  var happyLCount = 0;

  for (var i = 0; i < n - 1 ; i ++) {
    if (s[i] === "R" && s[i + 1] === "R") {
      happyRCount = happyRCount + 1;
    }
  }

  for (var i = n-1; i > 0 ; i --) {
    if (s[i] === "L" && s[i - 1] === "L") {
      happyLCount = happyLCount + 1;
    }
  }

  var happyCount = happyRCount + happyLCount;

  var maxHappyCount = 0;

  if (happyCount + k * 2 > n -1 ) {
    maxHappyCount = n -1;
  } else {
    maxHappyCount = happyCount + k * 2;
  }
  console.log(maxHappyCount);
}
  
Main(require("fs").readFileSync("/dev/stdin", "utf8"));