#1002

Easy

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Find Common Characters

Return an array of common characters from all strings in the given list of words, including duplicates.

Problem Statement

Given a list of strings, your task is to find and return an array of characters that appear in all strings, including duplicates. The order of the characters in the output can vary, and the solution should focus on efficiently identifying common characters through array scanning and hash lookups.

For example, with the input words = ["bella","label","roller"], the expected output is ["e","l","l"]. This method should be able to handle up to 100 strings, each containing up to 100 lowercase letters, within reasonable time limits.

Examples

Example 1

Input: words = ["bella","label","roller"]

Output: ["e","l","l"]

Example details omitted.

Example 2

Input: words = ["cool","lock","cook"]

Output: ["c","o"]

Example details omitted.

Constraints

1 <= words.length <= 100
1 <= words[i].length <= 100
words[i] consists of lowercase English letters.

Solution Approach

Array Scanning with Hash Table

Start by scanning through each string and using a hash table to store the count of each character. Then, iterate over the characters and check if the character count is consistent across all strings. This is an efficient way to track and identify common characters.

Minimizing Space Usage

To ensure space efficiency, you can minimize memory usage by reusing the hash table and updating it in place as you move through the words. This approach reduces the space complexity while ensuring that the character frequencies remain accurate.

Handling Duplicates

Pay attention to duplicates by tracking the minimum frequency of each character that appears across all words. If a character appears multiple times in some words, include it the same number of times in the result.

Complexity Analysis

Metric	Value
Time	O(n \cdot k)
Space	O(1)

The time complexity is O(n * k), where n is the number of strings and k is the length of the longest string, due to the need to scan each character in every word. Space complexity is O(1), as we use a constant amount of extra space for the hash table that stores character counts.

What Interviewers Usually Probe

Evaluate how well the candidate handles array scanning and hash lookups.
Watch for issues with efficiently tracking duplicates.
Look for the candidate’s ability to minimize space usage during implementation.

Common Pitfalls or Variants

Common pitfalls

Failing to account for duplicate characters and incorrectly handling them.
Using unnecessary extra space for storing counts that could be optimized.
Overcomplicating the solution instead of utilizing efficient hash lookups.

Follow-up variants

What if the list of words contains only one string?
What if there are no common characters between the words?
How would you modify the solution if the words can contain uppercase characters?

FAQ

How can I solve the Find Common Characters problem using hash tables?

By using hash tables, you can track the frequency of each character in every string, and then compare the counts across all words to find common characters.

What is the time complexity of the Find Common Characters problem?

The time complexity is O(n * k), where n is the number of words and k is the length of the longest word.

Are there any common pitfalls in solving this problem?

Yes, some common mistakes include failing to track duplicates correctly, using unnecessary space, and overcomplicating the solution with inefficient algorithms.

How do I handle duplicates in the Find Common Characters problem?

Track the minimum frequency of each character across all strings and include duplicates in the output based on this count.

Can I optimize space usage in the Find Common Characters problem?

Yes, by updating the hash table in place as you scan each word, you can minimize space usage while ensuring the accuracy of the character counts.

terminal

Solution

Solution 1: Counting

We use an array $cnt$ of length $26$ to record the minimum number of times each character appears in all strings. Finally, we traverse the $cnt$ array and add characters with a count greater than $0$ to the answer.

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class Solution:
    def commonChars(self, words: List[str]) -> List[str]:
        cnt = Counter(words[0])
        for w in words:
            t = Counter(w)
            for c in cnt:
                cnt[c] = min(cnt[c], t[c])
        return list(cnt.elements())

Continue Practicing

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