**Definitions**
Let $ n \in \mathbb{Z}^+ $ be an even integer denoting the number of digits in the ticket.
Let $ s \in \{0,1,\dots,9,\texttt{?}\}^n $ be the ticket string, where $ \texttt{?} $ denotes an erased digit.
Let $ q $ be the number of `?` characters in $ s $, with $ q $ even.
Partition $ s $ into two halves:
- Left half $ L = s[1:\frac{n}{2}] $, right half $ R = s[\frac{n}{2}+1:n] $.
Let $ S_L $ be the sum of known digits in $ L $, $ S_R $ the sum of known digits in $ R $.
Let $ q_L $ be the number of `?` in $ L $, $ q_R $ the number of `?` in $ R $, so $ q_L + q_R = q $.
**Constraints**
1. $ 2 \le n \le 2 \cdot 10^5 $
2. $ q $ is even, $ 0 \le q \le n $
3. Players alternate turns, Monocarp first. Each turn, a player replaces one `?` with a digit in $ \{0,1,\dots,9\} $.
4. Game ends when all `?` are filled.
**Objective**
Determine the winner under optimal play:
- Bicarp wins if $ S_L + \text{sum of digits added to } L = S_R + \text{sum of digits added to } R $.
- Monocarp wins otherwise.
Let $ d = S_R - S_L $ be the initial difference.
Let $ x $ be the total sum added to $ L $, $ y $ the total sum added to $ R $.
Bicarp wins iff $ x - y = d $.
Total moves: $ q $, Monocarp makes $ \frac{q}{2} $ moves, Bicarp makes $ \frac{q}{2} $ moves.
Monocarp controls $ \frac{q_L + q_R}{2} $ moves total, but can choose any `?` position.
Bicarp responds optimally.
**Key Insight**
The game reduces to:
- Monocarp aims to prevent $ x - y = d $.
- Bicarp aims to achieve $ x - y = d $.
- Each player chooses a position (left or right) and assigns a digit 0–9.
Define:
- $ a = q_L $, $ b = q_R $, so $ a + b = q $.
- Let $ m = \frac{q}{2} $: each player makes $ m $ moves.
Monocarp can choose to play on left or right; Bicarp responds.
Bicarp wins if and only if:
$$
d = \frac{9}{2}(b - a)
$$
and both players play optimally.
**Final Condition**
Bicarp wins if and only if:
$$
S_R - S_L = \frac{9}{2}(q_L - q_R)
$$
Otherwise, Monocarp wins.