This is the Catalan number problem in disguise, which I did eventually recognize, but not before spending a few minutes trying to brute-force small cases and pattern match.
Recognize that the problem is equivalent to counting Dyck paths of length 2n, where each buy is an up-step and each sell is a down-step, and the position never goes negative. The answer is the nth Catalan number, which can be computed efficiently using the formula C_n = (2n)!/(n!(n+1)!) or via dynamic programming with O(n^2) time or O(n) time using combinatorial formulas.
Pro tip: Mention that this is a classic problem with applications in many areas, and that the Catalan numbers grow exponentially, so for large n you might need to compute modulo a prime or use arbitrary precision arithmetic. Also, be prepared to discuss the time and space complexity of different computation methods.
Map each trade to a step: buying one share is an up-step, selling one share is a down-step. The condition that the position never goes negative means the path never goes below the starting level.
The sequences correspond to Dyck paths of length 2n, which are counted by the nth Catalan number. Explain why the bijection holds.
State the closed-form formula: C_n = (2n)!/(n!(n+1)!). Alternatively, mention the recurrence C_0=1, C_{n+1} = sum_{i=0}^n C_i C_{n-i}.
For small n, use the closed-form with factorials. For large n, use dynamic programming with O(n^2) time or compute using the recurrence with O(n) space. If needed, compute modulo a prime using modular inverses.
Mention that the number of sequences grows exponentially, so for large n the result may be huge. Discuss time and space complexity of the chosen method and potential overflow issues.
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