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 PyPy3 (2.4.0) (judge verdict: Time Limit Exceeded)

from collections import deque
N,K = (int(x) for x in input().split())
S = input()
count = []
d = deque(list(S))

now = d.popleft()
cum = 1
for i in range(len(S)-1):
    popped = d.popleft()
    if popped == now:
        cum += 1
    else:
        count.append(cum)
        now = popped
        cum = 1
count.append(cum)

if len(count) // 2 <= K:
    print(N-1)
else:
    m = 0
    for i in range(len(count)-2*K):
        if i == 0:
            c = sum(count[0:2*K+1]) - 1
            d = sum([j-1 for j in count[2*K+1:]])
        elif i == len(count) - 2*K + 1:
            c = sum(count[-(2*K+1):])
            d = sum([j-1 for j in count[0:-(2*K+1)]])
        else:
            c = sum(count[i:2*K+1+i]) - 1
            d1 = sum([j-1 for j in count[0:i]])
            d2 = sum([j-1 for j in count[2*K+1+i:]])
            d = d1 + d2
        m = max(m,c+d)
    print(m)