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 Python (3.4.3) (judge verdict: Wrong Answer)

N, K = map(int, input().split())
S = input()

left_ac = [0]
right_ac = [0]

prev = ""
for i in range(N):
    s = S[i]
    if s == "L":
        if i - 1 >= 0:
            if S[i - 1] == "L":
                left_ac.append(left_ac[-1] + 1)
            else:
                left_ac.append(left_ac[-1])
        else:
            left_ac.append(left_ac[-1])
        right_ac.append(right_ac[-1])
    else:
        if i + 1 < N:
            if S[i + 1] == "R":
                right_ac.append(right_ac[-1] + 1)
            else:
                right_ac.append(right_ac[-1])
        else:
            right_ac.append(right_ac[-1])
        left_ac.append(left_ac[-1])
left_ac.pop(0)
right_ac.pop(0)
# print(left_ac)
# print(right_ac)

left = 0
right = 0
ans = right_ac[-1] + left_ac[-1]
remain = K
while left < N:
    s = S[left]
    while remain > 0 and right < N:
        if S[right] != s:
            if (right + 1 < N and S[right + 1] == s) or (right == N - 1 and S[N - 1] != s):
                remain -= 1
        right += 1

    if s == "L":
        #print(s, right - left, left, right)
        tmp = right - left
        if right == N:
            tmp -= 1
        if right < N - 1:
            tmp += right_ac[-1] - right_ac[right + 1]
            tmp += left_ac[-1] - left_ac[right + 1]
        if left > 0:
            tmp += left_ac[left - 1] - left_ac[0]
            tmp += right_ac[left - 1] - right_ac[0]
        ans = max(ans, tmp)
    else:
        #print(s, right - left, left, right)
        tmp = right - left
        if right == N:
            tmp -= 1
        if right < N - 1:
            tmp += right_ac[-1] - right_ac[right+1]
            tmp += left_ac[-1] - left_ac[right+1]
        if left > 0:
            tmp += left_ac[left - 1] - left_ac[0]
            tmp += right_ac[left - 1] - right_ac[0]
        ans = max(ans, tmp)

    if right == left:
        right += 1
    if left + 1 < N and S[left + 1] != s:
        remain += 1
    left += 1
print(ans)