1190MediumArrayHash TableString

Two Barred Terms Hold the Ticket

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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.

A support ticket reaches the desk as a list of lowercase words, words, and the desk keeps a list of barred terms, barred.

The desk holds the ticket back once two of its words match a barred term exactly. Those two can be the same term matched twice, since they are still two words of the ticket.

Return true when the ticket is held back and false when it goes through.

Examples

Example 1

Input: words = ["refund", "late", "lost"], barred = ["refund", "lost"]
Output: true

Two of the three words of the ticket are barred terms, so the desk holds it.

Example 2

Input: words = ["refund", "refund"], barred = ["refund"]
Output: true

One barred term, but the ticket carries it twice, and two words of the ticket are what the desk counts.

Example 3

Input: words = ["carpet", "car"], barred = ["car", "van"]
Output: false

Only the second word matches a barred term. The first merely contains one, which is not a match, so one word is all the desk has.

Constraints

1 <= words.length <= 10^5
1 <= barred.length <= 10^5
1 <= words[i].length <= 15
1 <= barred[i].length <= 15
words[i] and barred[i] hold lowercase English letters 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 report_spam(words: list[str], barred: list[str]) -> bool:

Java

public boolean reportSpam(String[] words, String[] barred)

The intended approach

Time O(W + B) in the total letters of the two lists.Space O(B) for the set of barred terms.

The only question about a word of the ticket is whether it is one of the barred terms, so pour the barred list into a hash set once. Then walk the ticket, adding one to a running count for every word the set holds, and answer `true` as soon as that count reaches two. Counting words rather than distinct terms is what makes a term matched twice enough on its own, so nothing needs remembering about which term was matched. The walk can stop the moment the count reaches two, though even a full walk stays linear in the number of letters handed over.

Pythonentry point: report_spam

def report_spam(words, barred):
    blocked = set(barred)
    hits = 0
    for word in words:
        if word in blocked:
            hits += 1
            if hits == 2:
                return True
    return False

The obvious version

Compare every pair by hand. For each word of the ticket, walk the whole barred list looking for an exact match and add one to a tally when it turns up, then answer on whether the tally reached two. Costs a pass over the barred list for every word of the ticket.

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_report_spam(words, barred):
    hits = 0
    for word in words:
        for term in barred:
            if word == term:
                hits += 1
                break
    return hits >= 2

How this problem was checked

Reference solution executed against 15 test cases — 15 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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