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.

#1002

Easy

auto_awesome数组·哈希·扫描

LeetCode 题解工作台

查找共用字符

给你一个字符串数组 words ，请你找出所有在 words 的每个字符串中都出现的共用字符（包括重复字符），并以数组形式返回。你可以按任意顺序返回答案。示例 1：输入： words = ["bella","label","roller"] 输出： ["e","l","l"] 示例 2：…

数组哈希表字符串

题目描述

给你一个字符串数组 words ，请你找出所有在 words 的每个字符串中都出现的共用字符（包括重复字符），并以数组形式返回。你可以按 任意顺序 返回答案。

示例 1：

输入：words = ["bella","label","roller"]
输出：["e","l","l"]

示例 2：

输入：words = ["cool","lock","cook"]
输出：["c","o"]

提示：

1 <= words.length <= 100
1 <= words[i].length <= 100
words[i] 由小写英文字母组成

lightbulb

解题思路

方法一：计数

我们用一个长度为 $26$ 的数组 $cnt$ 记录每个字符在所有字符串中出现的最小次数，最后遍历 $cnt$ 数组，将出现次数大于 $0$ 的字符加入答案即可。

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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())

speed

复杂度分析

指标	值
时间	O(n \cdot k)
空间	O(1)

psychology

面试官常问的追问

外企场景

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

warning

常见陷阱

外企场景

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

swap_horiz

进阶变体

外企场景

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

help

常见问题

外企场景

继续练习

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