0191HardStringDynamic ProgrammingGreedyRecursion

Dispatch Rule Match

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Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

An event router picks a destination by testing an incoming event's key against a subscription rule.

The key is plain text. The rule is plain text too, except that it may carry two wildcard symbols:

? stands in for exactly one character of the key.
* stands in for a stretch of the key of any length, and a stretch of length zero is allowed.

The router fires only on a total fit: reading the rule left to right must consume the key from its first character through its last, with nothing of the key spilling past the end of the rule and nothing of the rule left dangling once the key is used up.

Return true when rule fires on key, and false when it does not. Either text may be empty.

Examples

Example 1

Input: key = "deploy", rule = "d*y"
Output: true

The wildcard soaks up `eplo`, leaving `d` and `y` to line up with the two ends of the key.

Example 2

Input: key = "deploy", rule = "d?ploy?"
Output: false

The first `?` takes `e` and the literals take `ploy`, but the trailing `?` still demands a character and the key has run dry.

Example 3

Input: key = "", rule = "***"
Output: true

Each wildcard is free to stand in for a stretch of length zero, so an empty key is a total fit.

Constraints

0 <= key.length, rule.length <= 2000
key holds lowercase letters only
rule holds lowercase letters, '?' and '*' only

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def rule_fires(key: str, rule: str) -> bool:

Java

public boolean ruleFires(String key, String rule)

The intended approach

Time O(n*m) worst caseSpace O(1)

Walk both texts with one cursor each. Literals and `?` advance both cursors in lockstep. On a `*` remember where it sat and how far the key had been read, then let it stand for nothing at first; whenever a later character refuses to line up, hand one more key character to that remembered wildcard and retry from just after it. Only the most recent wildcard ever needs to grow, since an earlier one could always have taken the same characters.

Pythonentry point: rule_fires

def rule_fires(key, rule):
    # Time O(n*m) in the worst case: each retry after a wildcard rescans
    # at most the rule, and the key cursor never moves backwards past a retry.
    # Space O(1): four integer cursors, no table.
    i = j = 0
    star = -1
    resume = 0
    while i < len(key):
        if j < len(rule) and (rule[j] == '?' or rule[j] == key[i]):
            i += 1
            j += 1
        elif j < len(rule) and rule[j] == '*':
            star = j
            resume = i
            j += 1
        elif star != -1:
            resume += 1
            i = resume
            j = star + 1
        else:
            return False
    while j < len(rule) and rule[j] == '*':
        j += 1
    return j == len(rule)

The obvious version

Plain recursion with no memo: peel the rule one symbol at a time and, for a wildcard, try both standing for nothing and swallowing one more character. Exponential.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_rule_fires(key, rule):
    if not rule:
        return not key
    head = rule[0]
    if head == '*':
        if bf_rule_fires(key, rule[1:]):
            return True
        if not key:
            return False
        return bf_rule_fires(key[1:], rule)
    if not key:
        return False
    if head == '?' or head == key[0]:
        return bf_rule_fires(key[1:], rule[1:])
    return False

How this problem was checked

Reference solution executed against 21 test cases — 21 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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