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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define REP(i, n) for(int i = 0; i < (int)n; i++)
int main()
{
  bool DEBUG = true;
  int n, k;
  cin >> n >>k;
  string s;
  cin >> s;
  int ans = 0;
  if(s[0] == 'R' && s[1] == 'R'){ans ++;}
  if(s[n - 1] == 'L' && s[n - 2] == 'L'){ans ++;}
  for (int i = 1; i < n - 1;i ++){
    if(s[i] == 'R'&&s[i + 1] == 'R'){
      ans ++;
    }
    if(s[i] == 'L'&&s[i - 1] == 'L'){      
      ans ++;
    }
  }
  if(DEBUG)cout << ans << endl;
  int numlr = 0;
  int numrl = 0;
  // if(s[0] == 'L')numrl++;
  // if(s[n - 1] == 'R')numlr++;
    // cout << s << endl;
  for (int i = 1; i < n - 1;i ++){
    if(s[i] == 'R'&&s[i + 1] == 'L') {
      numrl ++;
      // cout << i << endl;
    }
    if(s[i] == 'L'&&s[i + 1] == 'R') {
      numlr ++;
      // cout << i << endl;
    }
  }
  // if(DEBUG) cout << numrl << endl;
  // if(DEBUG) cout << numlr << endl;
  if(k <= min(numrl, numlr)){
    cout << ans+2*min(k,min(numrl, numlr)) << endl;
  }else{
    if(s[0] == 'L')numrl++;
    if(s[n - 1] == 'R')numlr++;
    int mini = min(numrl, numlr);
    // cout << mini << endl;x
    ans += 2*(k - mini);
    cout << ans + min(min((k + 1- mini), 2),mini) << endl;
  }

}