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: Accepted)


import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.NoSuchElementException;
import java.util.stream.IntStream;

public class Main {

	public static void main(String[] args) throws IOException {
		new Main().solve();
	}

	private void solve() throws IOException {
		try {
			//			solveA();
			//			solveB();
			//			solveC();
			solveD();
			//			 solveE();
			//			 solveF();
		} finally {
			if (in != null) {
				in.close();
			}
			if (out != null) {
				out.flush();
				out.close();
			}
		}

	}

	private void solveA() {
		int n = nextInt();

		out.println((int) Math.pow(n, 3));
	}

	private void solveB() {
		int n = nextInt();
		int[] a = IntStream.range(0, n).map(i -> nextInt()).toArray();
		int[] b = IntStream.range(0, n).map(i -> nextInt()).toArray();
		int[] c = IntStream.range(0, n - 1).map(i -> nextInt()).toArray();

		int res = 0;
		for (int i = 0; i < n; i++) {
			int curr = a[i] - 1;
			res += b[curr];
			if (i < n - 1 && a[i] + 1 == a[i + 1])
				res += c[curr];
		}

		out.println(res);
	}

	private void solveC() {
		int n = nextInt();
		int[] wk = IntStream.range(0, n - 1).map(i -> nextInt()).toArray();
		int[] res = new int[n];
		res[n - 1] = wk[n - 2];
		for (int i = n - 2; i >= 0; i--) {
			if (i != 0) {
				res[i] = Math.min(wk[i], wk[i - 1]);
			} else {
				res[i] = wk[i];
			}
		}
		long total = Arrays.stream(res).sum();
		out.println(total);
	}

	private void solveD() {
		int n = nextInt();
		int k = nextInt();
		char[] s = next().toCharArray();
		List<char[]> run = new ArrayList<char[]>();
		char pre = s[0];
		int cnt = 1;
		int happy = 0;
		for (int i = 1; i < s.length; i++) {
			if (pre == s[i]) {
				cnt++;
			} else {
				run.add(new char[] { pre, (char) ('0' + cnt) });
				happy += (cnt - 1);
				pre = s[i];
				cnt = 1;
			}
		}
		run.add(new char[] { pre, (char) ('0' + cnt) });
		happy += (cnt - 1);

		int runLengthElm = run.size();
		/*
		 * 要素の数の偶奇で場合分け
		 */
		if (runLengthElm % 2 == 1) {
			/*
			 * 奇数の場合は、両端が同じ方向を向いている
			 * そのため、全量動かせばよい
			 */
			int switchCnt = runLengthElm / 2;
			int count = (k - switchCnt) > 0 ? switchCnt * 2 : k * 2;
			out.println(happy + count);
		} else {
			/*
			 * 偶数の場合は、両端が違う方向を向いている
			 * また、switchしている個所は奇数となる
			 */
			int switchCnt = (runLengthElm - 1) / 2;
			int count = (k - switchCnt) > 0 ? switchCnt * 2 : k * 2;
			if (switchCnt > k) {
				/*
				 * switch箇所がKより大きいならとりあえず動かせるだけ動かせばよい
				 */
				k = 0;
			} else {
				/*
				 * switch箇所がKより小さい場合、
				 */
				if(k != switchCnt)
					count++;
//				k -= switchCnt;
			}
//			if (k > 0) {
			//			}
			out.println(happy + count);
		}

	}

	private void solveE() {
		int n = nextInt();

		out.println("");
	}

	private void solveF() {
		int n = nextInt();

		out.println("");
	}

	/*
	 * 階乗計算の最大値
	 * もっと大きいのが必要ならここを修正
	 */
	private final int FACTORIAL_NUM = 111111;
	/*
	 * 階乗の事前計算（mod 済）
	 */
	private long[] factorial = new long[FACTORIAL_NUM];
	/*
	 * 逆数の事前計算（mod 済）
	 */
	private long[] reverseFactorial = new long[FACTORIAL_NUM];
	private boolean isInitFactorial = false;

	/**
	 * 階乗計算の事前累積和
	 * [1, 1, 2, 3, 4, 5, … FACTORIAL_NUM]
	 * mod済
	 */
	private void initFactorial(long mod) {
		factorial[0] = 1;
		for (int i = 1; i < FACTORIAL_NUM; i++) {
			factorial[i] = (factorial[i - 1] * i) % mod;
		}
	}

	/**
	 * 逆数の事前計算
	 * もっと速い方法があったので修正
	 */
	private void initReverseFactorial(long mod) {

		reverseFactorial[FACTORIAL_NUM - 1] = modInverse(factorial[FACTORIAL_NUM - 1], mod);
		for (int i = FACTORIAL_NUM - 2; i >= 0; i++) {
			/*
			 * 以下を利用する。
			 * これにより毎回modInverseしなくてよくなる
			 * (N!)^(-1) * N -> ((N - 1)!)^(-1)
			 * (N-1!)^(-1) * N -> ((N - 2)!)^(-1)
			 */
			reverseFactorial[i] = (reverseFactorial[i + 1] * (i + 1)) % mod;
		}
		//		reverseFactorial[0] = 1;
		//		for (int i = 0; i < FACTORIAL_NUM; i++) {
		//			reverseFactorial[i] = modInverse(factorial[i], mod);
		//		}
	}

	/**
	 * フェルマーの小定理を用いて逆元を求める。
	 * ある数xのmod p（pは素数）の逆数x'はx' = x^(p-2)
	 * 繰り返し二乗法を用いて計算する。
	 * http://satanic0258.hatenablog.com/entry/2016/04/29/004730
	 * {@link BigInteger#modInverse(BigInteger)}とどちらが速いか？
	 * @param x
	 * @return
	 */
	private long modInverse(long x, long mod) {
		long res = 1L;
		long k = mod - 2L;
		long y = x;
		while (k != 0) {
			if (k % 2 != 0) {
				res = (res * y) % mod;
			}
			y = (y * y) % mod;
			k /= 2;
		}
		return res;
	}

	private void initFactorialCalc(long mod) {
		if (isInitFactorial) {
			//初期化済みならreturn
			return;
		}
		//階乗の事前計算
		initFactorial(mod);
		//逆数の計算もしておく
		initReverseFactorial(mod);
		isInitFactorial = true;
	}

	/**
	 * 組み合わせ
	 * nCk -> n! / k!・(n-k)!
	 * @param n
	 * @param k
	 * @param mod
	 * @return
	 */
	private long combination(int n, int k, long mod) {
		initFactorialCalc(mod);
		if (k < 0 || n < k) {
			return 0;
		}

		//分子の計算
		long calResN = factorial[n];

		//分母の計算
		long calResNK = reverseFactorial[n - k];
		long calcResK = reverseFactorial[k];
		long denominator = (calResNK * calcResK) % mod;

		return (calResN * denominator) % mod;

	}

	/**
	 * 順列
	 * nPk -> n! / (n-k)!
	 * @param n
	 * @param k
	 * @return
	 */
	private long permutation(int n, int k, long mod) {
		initFactorialCalc(mod);
		if (k < 0 || n < k) {
			return 0;
		}
		long calResN = factorial[n];
		long calResNK = reverseFactorial[n - k];
		return (calResN * calResNK) % mod;
	}

	/**
	 * 多項係数
	 * 文字aをp個,bをq個,cをr個, dをs個 あわせてn個を1列に並べるときの並べ方
	 * n! / p!・q!・r!・s!
	 * @param n
	 * @param strNum
	 * @param mod
	 * @return
	 */
	private long multinominalCoefficient(int n, List<Integer> strNum, int mod) {
		initFactorialCalc(mod);

		long calResN = factorial[n];
		long calResDenomi = 1;
		for (Integer num : strNum) {
			calResDenomi *= reverseFactorial[num];
		}
		return (calResN * calResDenomi) % mod;
	}

	/**
	 * 重複組合せ nHr （同次積）
	 * nHr = (n+r-1)Cr
	 * 異なるn個のものから重複を許してr個取る組合せの総数
	 * 例：
	 * リンゴ，ミカン，牛肉の3種類の果物があります．これらの中から6個の食材を買って帰ります．
	 * このとき，何通りの買い方がありますか？ただし，含まれない食材があってもよいものとします
	 * @param n
	 * @param r
	 * @param mod
	 * @return
	 */
	private long homogeneousProduct(int n, int r, long mod) {
		if (n == 0 && r == 0) {
			return 1;
		}
		return combination(n + r - 1, r, mod);
	}

	private final PrintWriter out = new PrintWriter(System.out);
	private final InputStream in = System.in;
	private final byte[] buffer = new byte[1024];
	private int ptr = 0;
	private int buflen = 0;

	private boolean hasNextByte() {
		if (ptr < buflen) {
			return true;
		} else {
			ptr = 0;
			try {
				buflen = in.read(buffer);
			} catch (IOException e) {
				e.printStackTrace();
			}
			if (buflen <= 0) {
				return false;
			}
		}
		return true;
	}

	private int readByte() {
		if (hasNextByte())
			return buffer[ptr++];
		else
			return -1;
	}

	private static boolean isPrintableChar(int c) {
		return 33 <= c && c <= 126;
	}

	private void skipUnprintable() {
		while (hasNextByte() && !isPrintableChar(buffer[ptr]))
			ptr++;
	}

	public boolean hasNext() {
		skipUnprintable();
		return hasNextByte();
	}

	public int nextInt() {
		return Integer.parseInt(next());
	}

	public String next() {
		if (!hasNext())
			throw new NoSuchElementException();
		StringBuilder sb = new StringBuilder();
		int b = readByte();
		while (isPrintableChar(b)) {
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}

	public long nextLong() {
		if (!hasNext())
			throw new NoSuchElementException();
		long n = 0;
		boolean minus = false;
		int b = readByte();
		if (b == '-') {
			minus = true;
			b = readByte();
		}
		if (b < '0' || '9' < b) {
			throw new NumberFormatException();
		}
		while (true) {
			if ('0' <= b && b <= '9') {
				n *= 10;
				n += b - '0';
			} else if (b == -1 || !isPrintableChar(b)) {
				return minus ? -n : n;
			} else {
				throw new NumberFormatException();
			}
			b = readByte();
		}
	}
}