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NeetCode #133LC-884EasyArrays & Hashing
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#133 · #884 · Uncommon Words from Two Sentences(两句话中的不常见单词)

📌 Problem Statement & Constraints

A word is uncommon if it appears exactly once in either of the two sentences and does not appear in the other. Given two sentences s1 and s2, return a list of all uncommon words. Constraints: 1 <= s1.length, s2.length <= 200; sentences contain lowercase letters and single spaces.

💡 Core Algorithmic Approaches

  1. Concatenate the two sentences and count the frequency of every word.
  2. Words occurring exactly once across the combined text are precisely the uncommon ones -- a word appearing in both sentences would have a count of at least 2.
  3. Collect the words with count 1.
  4. This single-counter approach elegantly avoids tracking the two sentences separately.

💻 Benchmark Python3 Implementation

class Solution:
    def uncommonFromSentences(self, s1: str, s2: str) -> List[str]:
        from collections import Counter
        cnt = Counter((s1 + " " + s2).split())
        return [w for w, c in cnt.items() if c == 1]   # count 1 in the union

⚡ Complexity Deep Dive

⏱️ Time Complexity
O(n + m): splitting and counting are linear in the total text length.
💾 Space Complexity
O(n + m) for the counter.

⚠️ Interview Pitfalls & Follow-ups

  • Building two separate counters and taking the difference: it works, but you would still need to filter words with count 1 within each sentence; the combined counter does both jobs at once.
  • Forgetting the separating space when concatenating: "ab" + "cd" would merge two words into one. Use s1 + " " + s2.
  • Using set(s1.split()) ^ set(s2.split()): this loses multiplicity, so a word appearing twice in s1 and not at all in s2 would be wrongly included.
  • Assuming words are unique within a sentence: they are not, which is exactly what the count-1 filter handles.