1002. Find Common Characters

Problem¶

Return all lowercase letters that appear in every word, keeping duplicates.

Example:

words = ["bella", "label", "roller"]
common counts:
- e -> min(1,1,1) = 1
- l -> min(2,2,2) = 2
answer = ["e", "l", "l"]

Key trick¶

Count letters in each word, then keep the minimum frequency per letter across all words.

Trap¶

• Forgetting duplicates matter.
• Using set intersection, which loses counts.
• Typo: Counters does not exist, it is Counter.
• Mutating the first counter is fine, but be clear about it.

Why is it interesting?¶

It is a small problem that tests whether you see the difference between:

• common distinct letters
• common letters with multiplicity

It is also a clean use of frequency counting and intersection.

Python solution¶

from collections import Counter

class Solution:
    def commonChars(self, words: list[str]) -> list[str]:
        common = Counter(words[0])
        for word in words[1:]:
            common &= Counter(word)
        return list(common.elements())

Comment on my solution¶

• The idea is correct: use Counter for each word, then intersect them.
• from collections import Counters is a bug; it must be Counter.
• You do not need to store all counters first; intersect on the fly.
• count is a bit vague; common is clearer.

from collections import Counter

class Solution:
    def commonChars(self, words: list[str]) -> list[str]:
        counts = []
        for w in words:
            counts.append(Counter(w))
        count = counts[0]
        for c in counts[1:]:
            count = count & c
        return list(count.elements())