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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<cstdio>
#include<algorithm>
using namespace std;

int main(){
	int N, K;
	scanf("%d %d\n", &N, &K);
	bool is_L[N + 2];
	is_L[0] = false;
	is_L[N + 1] = true;
	for (int i = 1; i <= N; ++i){
		char s;
		scanf("%c", &s);
		is_L[i] = s == 'L';
	}
	int happy = 0;
	for (int i = 1; i <= N; ++i){
		if(is_L[i - 1] && is_L[i]) happy++;
		else if(!is_L[i] && !is_L[i + 1]) happy++;
	}
	int lr = 0, rl = 0;
	for (int i = 2; i <= N; ++i){
		if(is_L[i - 1] && !is_L[i]) lr++;
		else if(!is_L[i - 1] && is_L[i]) rl++;
	}
	int psuedo_rl = 0;
	if(is_L[1]) psuedo_rl++;
	if(!is_L[N]) psuedo_rl++;
	int manipulation = min({lr, rl, K});
	happy += 2*manipulation;
	lr -= manipulation;
	rl -= manipulation;
	K -= manipulation;
	manipulation = 0;
	if(is_L[1]){
		if(lr > rl){
			manipulation = rl*2 + 1;
			lr -= rl + 1;
			rl = 0;
			is_L[1] = false;
		}else{
			manipulation = lr*2;
			lr = 0;
			rl -= lr;
		}
		if(manipulation >= K){
			printf("%d\n", happy + K);
			return 0;
		}
		happy += manipulation;
		K -= manipulation;
	}else{
		if(rl > lr){
			manipulation = lr*2 + 1;
			rl -= lr + 1;
			lr = 0;
		}else{
			manipulation = rl*2;
			rl = 0;
			lr -= rl;
		}
		if(manipulation >= K){
			printf("%d\n", happy + K);
			return 0;
		}
		happy += manipulation;
		K -= manipulation;		
	}
	if(!is_L[N]){
		if(lr > rl){
			manipulation = rl*2 + 1;
			lr -= rl + 1;
			rl = 0;
		}else{
			manipulation = lr*2;
			lr = 0;
			rl -= lr;
		}
		if(manipulation >= K){
			printf("%d\n", happy + K);
			return 0;
		}
		happy += manipulation;
		K -= manipulation;
	}else{
		if(rl > lr){
			manipulation = lr*2 + 1;
			rl -= lr + 1;
			lr = 0;
		}else{
			manipulation = rl*2;
			rl = 0;
			lr -= rl;
		}
		if(manipulation >= K){
			printf("%d\n", happy + K);
			return 0;
		}
		happy += manipulation;
		K -= manipulation;		
	}
	printf("%d\n", happy);
	return 0;
}