Passing the Ball

Research notes: I checked the official Codeforces editorial, which points to the compact RL-position solution, and CF Step, which shows the direct simulation idea. The supplied statement is still the source of truth; these just confirm the intended direction.

Let positions be $1$-indexed. Define

$$k = \min\{i \mid s_i = \texttt{L}\}.$$

This $k$ always exists because the last character is L, and $k \ge 2$ because the first character is R.

For every position $i < k$, we have $s_i = \texttt{R}$ by definition of the first L. So the ball starts at student $1$, then goes

$$1 \to 2 \to 3 \to \cdots \to k.$$

That already visits exactly the students $1,2,\ldots,k$.

Now look at what happens at student $k$. Since $s_k = \texttt{L}$, student $k$ passes the ball to $k-1$. But $k$ was the first L, so $s_{k-1}=\texttt{R}$, meaning student $k-1$ passes the ball right back to $k$.

So after reaching $k$, the ball gets stuck in the 2-cycle

$$k-1 \leftrightarrow k.$$

No later student can ever be reached. They might have all kinds of spicy letters over there, but the ball does not care. It is trapped.

Because $k \le n$, the process has enough passes to reach student $k$ before it ends. Therefore the set of students who receive the ball is exactly

$$\{1,2,\ldots,k\},$$

so the answer is $k$.

In C++ zero-based indexing, the first L is at index $k-1$. Equivalently, the first substring RL starts at index $k-2$, so

$$\text{answer} = \texttt{s.find("RL")} + 2.$$

Either s.find('L') + 1 or s.find("RL") + 2 works. We use the latter since it matches the intended observation cleanly.

Edge cases:

• If $n=2$, then $s=\texttt{RL}$ and the answer is $2$.
• If $s=\texttt{RRR...RL}$, then $k=n$, so every student receives the ball.
• Characters after the first L are irrelevant because the ball never reaches them.

Complexity per test case is $O(n)$ time and $O(1)$ extra memory. With $n \le 50$, this is tiny. Like, not even worth waking the CPU up for.