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NeetCode #43LC-929EasyArrays & Hashing
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#43 · #929 · Unique Email Addresses(独特的电子邮件地址)

📌 Problem Statement & Constraints

Every valid email address consists of a local name, an @, and a domain name. In the local name, periods are ignored and everything from a + onward is ignored. Given a list of emails, return the number of distinct addresses that actually receive mail. Constraints: 1 <= emails.length <= 100.

💡 Core Algorithmic Approaches

  1. Split each address at the @ to separate the local name from the domain.
  2. Normalise the local name: cut it at the first +, then remove all . characters.
  3. Reassemble as normalised_local + '@' + domain and add it to a set.
  4. The answer is the size of the set. The + truncation must happen before the dot removal (or vice versa) -- either order works, but the + cut must come first if you split on +, since the domain never contains a +.

💻 Benchmark Python3 Implementation

class Solution:
    def numUniqueEmails(self, emails: List[str]) -> int:
        seen = set()
        for e in emails:
            local, domain = e.split("@")
            local = local.split("+")[0]     # drop everything after '+'
            local = local.replace(".", "")  # dots are ignored
            seen.add(local + "@" + domain)
        return len(seen)

⚡ Complexity Deep Dive

⏱️ Time Complexity
O(total characters): each email is normalised in one pass.
💾 Space Complexity
O(n) for the set of distinct addresses.

⚠️ Interview Pitfalls & Follow-ups

  • Applying the dot/plus rules to the domain: they apply only to the local name; [email protected] and [email protected] are the same mailbox, but [email protected] is not the same as [email protected].
  • Splitting on + without limiting the split: local.split('+')[0] is correct; local.split('+') and taking a fixed index is not.
  • Using replace('.', '') on the whole email: this would also strip dots from the domain.
  • Forgetting that emails are guaranteed valid: no need for format validation.