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easy +15 pts

Valid parentheses

Stack intuition in nine lines.

Implement `is_valid_paren(s: str) -> bool` for characters in `()[]{}`. Every opener must close without crossing.

Constraints

0 ≤ len(s) ≤ 10^4; only the six delimiter characters.

Example

>>> is_valid_paren("([{}])")
True
>>> is_valid_paren("(]")
False
15 points ~14 min

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Push openers; on closers, check stack top matches.
Python 3
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