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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>

char s[100001];

void swap(int l, int r) {
   char c,d;

   while (l<=r) {
      if (s[l]=='L') c='R';
      else           c='L';
      if (s[r]=='L') d='R';
      else           d='L';
      s[l]=d;
      s[r]=c;
      l++;r--;
   }
}

int main(){
   int n,k,i,j,l,a;
 
   scanf("%d %d",&n,&k);
   scanf("%s",s);
   for (j=i=0;i<k;i++) {
      for (;j<n-2;j++)
         if (s[j]=='L' && s[j+1]=='R') {
            for (l=j+1;l<n-1;l++)
               if (s[l]=='R' && s[l+1]=='L') break;
            if (l==n-1) continue;
            swap(j+1,l);
            break;
         } else if (s[j]=='R' && s[j+1]=='L') {
            for (l=j+1;l<n-1;l++)
               if (s[l]=='L' && s[l+1]=='R') break;
            if (l==n-1) continue;
            swap(j+1,l);
            break;
         }
      if (j==n-2) break;
   }
   if (i<k) {
      for (i=1;i<n;i++)
         if (s[0]!=s[i]) break;
      if (i<n) {
         j=i;
         for (;i<n;i++)
            if (s[0]==s[i]) break;
         if (i==n) swap(j,n-1);
      }
   }
   for (a=i=0;i<n;i++)
      if (s[i]=='L' && i>0 && s[i-1]=='L') a++;
      else if (s[i]=='R' && i<n-1 && s[i+1]=='R') a++;
   printf("%d",a);
   return 0;
}