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 9.2.1) (judge verdict: Wrong Answer)

#include<iostream>
using namespace std;
int N,K;
int calc(int *a){
    int out=0;
    int temp=0;
    for(int i=0;i<N;i++){
        if(a[i])
            temp=0;
        else{
            out+=temp;
            temp=1;
        }
    }
    temp=0;
    for(int i=N-1;i>=0;i--){
        if(a[i]){
            out+=temp;
            temp=1;
        }
        else{
            temp=0;
        }
    }
    return out;
}
void change(int *a,int i,int j){
    for(;i<=j;i++,j--){
        if(!(a[i]^a[j])){
            a[i]=a[j]=!a[j];
        }
    }
    for(i=0;i<N;i++){
        cout<<a[i]<<" ";
    }
}
void scan(int* a,int *out){
    int m=0,n=0;
    for(int i=1;i<N-1;i++){
        int _break=0;
        if(a[i-1]^a[i]){
            for(int j=i;j<N-1;j++){
                if(a[j]^a[j+1]&&(a[j]==a[i])){
                    m=i,n=j;
                    *out+=2;
                    _break=1;
                    break;
                }
            }
        }
        if(_break)
            break;
    }
    if(!m){
        for(int i=1;i<N-1;i++){
            if(a[i]^a[i+1]&&(a[i]==a[0])){
                n=i;
                *out+=1;
                break;
            }
        }
    }
    if(!n){
        for(int i=1;i<N-1;i++){
            if(a[i]^a[i-1]&&(a[i]==a[N-1])){
                m=i;
                n=N-1;
                *out+=1;
                break;
            }
        }
    }
    change(a,m,n);
}
int main(){
    cin>>N>>K;
    int* a=new int[N];
    char S;
    for(int i=0; i<N; i++){
        cin>>S;
        if(S=='L')
            a[i]=0;
        else
            a[i]=1;
    }
    int out=calc(a);
    for(int k=0;k<K;k++){
        scan(a,&out);
    }
    cout<<out<<endl;
    return 0;
}