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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)


import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collection;
import java.util.Collections;
import java.util.HashMap;
import java.util.Scanner;
import java.util.HashSet;
import java.util.Iterator;

class Main {

    static final int MOD = 1000000007;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        final int N = sc.nextInt();
        final int K = sc.nextInt();
        String S = sc.next();
        char[] ch = S.toCharArray();
        boolean[] b = new boolean[N];
        int lc = 0;
        int rc = 0;
        ArrayList<Integer> LRp = new ArrayList<>();
        ArrayList<Integer> RLp = new ArrayList<>();
        ArrayList<Integer> P = new ArrayList<>();
        int cnt = 0;
        for (int i = 0; i < N; i++) {
            b[i] = ch[i] == 'L' ? true : false;
        }
        int ans = 0;
        for (int i = 0; i < N - 2; i++) {
            if (b[i] && !b[i + 1]) {
                LRp.add(i);
                P.add(i);
            } else if (!b[i] && b[i + 1]) {
                RLp.add(i);
                P.add(i);
            }
        }
        P.add(N - 1);
        for (int i = Math.max(0,Math.min(1,P.size()-2)); i < P.size() - 1 && cnt < K; i+=2) {
            for (int k = P.get(i)+1; k <= P.get(i + 1); k++) {
                b[k] = b[k - 1];
            }
            cnt++;
        }
        for (int i = 1; i < N; i++) {
            if (b[i] && b[i - 1])
                ans++;
        }
        for (int i = N - 2; i >= 0; --i) {
            if (!b[i] && !b[i + 1])
                ans++;
        }
        // ans += Math.min(LRp.size(), K) * 2;
        // if (b[0] != b[N - 1] && LRp.size() > RLp.size()) {
        //     ans--;
        // }
        // System.out.println(LRp.size() + " " + RLp.size());
        // System.out.println(cnt + Arrays.toString(b));
        System.out.println(ans);
    }
}