LeetCode: Count Common Words With One Occurrence

Problem:

https://leetcode.com/problems/count-common-words-with-one-occurrence/description/

Code:

class Solution:
    def countWords(self, words1: List[str], words2: List[str]) -> int:
        frequency1 = {}
        frequency2 = {}
        count=0

        for word in words1:
            frequency1[word] = frequency1.get(word, 0) + 1

        for word in words2:
            frequency2[word] = frequency2.get(word, 0) + 1

        for word in frequency1.keys():
            if frequency1.get(word, 0) == 1 and frequency2.get(word, 0) == 1:
                count+=1

Key Points:

• Goal: Count how many words appear: • exactly once in words1 AND • exactly once in words2.

• Build two frequency maps: frequency1[word] = how many times word appears in words1 frequency2[word] = how many times word appears in words2

• Why use frequency1.get(word, 0)?

  • If the word does NOT exist in the dictionary, .get() safely returns 0 instead of causing a KeyError.

  • This makes checks like (frequency2.get(word, 0) == 1) safe and clean without needing "if word in dictionary".

  • It simplifies code and avoids bugs.

• After building both maps: Loop through each unique word in words1: - If frequency1[word] == 1 AND frequency2.get(word, 0) == 1: → this word appears once in both lists → increase the result count

• Why loop through frequency1.keys() instead of words2?

  • Because a valid word MUST appear in words1 exactly once.

  • No need to check words that don't appear in words1 at all.

• This method avoids:

  • Nested loops

  • Repeated scanning of lists

  • Extra sets or sorting

• Time Complexity: O(n + m)
  (Single pass to count, single pass to compare)

• Space Complexity: O(n + m)
  (Two frequency maps)

• Tools used:

  • Hashmaps (dicts) → fast O(1) frequency lookup

  • .get(key, default) → safe access for missing words