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<iostream>
#include<algorithm>
#include<bitset>
#include<cmath>
#include<complex>
#include<deque>
#include<functional>
#include<iomanip>
#include<iostream>
#include<iterator>
#include<map>
#include<numeric>
#include<queue>
#include<set>
#include<stack>
#include<string>
#include<unordered_map>
#include<unordered_set>
#include<utility>
#include<vector>
#include<limits.h>

using namespace std;
typedef long long ll;
typedef pair<int,int> PI;
typedef pair<int,pair<int,int>> PII;
static const int IINF=INT32_MAX;
static const ll LINF=INT64_MAX;
static const ll mod=1e9+7;
static const int dx[4]={1,-1,0,0};
static const int dy[4]={0,0,1,-1};

template<class T> inline bool chmin(T& a,T b){if(a>b){a=b;return true;}return false;}
template<class T> inline bool chmax(T& a,T b){if(a<b){a=b;return true;}return false;}

int N,K;
string s;

int calc(int sl,string S){
    int t=0;
    int l=sl,r=-1;
    char cl;
    if(l==0)cl=S[l];
    int res=0;
    for(int i=1;i<N-1;++i){
        if(t>=K)break;
        if(l==-1){
            if(S[i]!=S[i-1]){
                l=i;
                cl=S[i];
            }
        }
        else if(l!=-1){
            if(S[i]!=S[i+1]&&S[i]!=cl){
                S[l]=S[i];
                S[i]=cl;
                if(l==0)res++;
                else res+=2;
                l=-1;
                r=i+1;
                t++;
                i--;
                //cout<<S<<endl;
            }
        }
    }
    //cout<<cl<<endl;
    //cout<<l<<endl;
    //cout<<res<<endl;
    if(l!=-1&&cl!=S[N-1]&&t<K)res++;
    return res;
}

int main(){
    cin>>N>>K;
    cin>>s;
    int ans=0;
    for(int i=0;i<N;++i){
        if(i!=0){
            if(s[i]=='L'&&s[i-1]=='L')ans++;
        }
        if(i!=N-1){
            if(s[i]=='R'&&s[i+1]=='R')ans++;
        }
    }
    ans+=max(calc(-1,s),calc(0,s));
    cout<<min(ans,N-1)<<endl;
}