What is the best approach for grouping anagrams in terms of performance?

Using hash tables with either a sorted key or a frequency count of characters provides the most efficient grouping. The frequency count approach is usually better for performance when avoiding unnecessary sorting.

How do I handle edge cases like empty strings in this problem?

Empty strings should be treated like any other string and grouped in their own group. Make sure to include logic that correctly handles this case.

Is sorting the strings the only way to group anagrams?

No, another approach is to use a frequency count of characters as a key for the hash table. Sorting is not necessary and can be more expensive in terms of time complexity.

How does the time complexity differ between sorting and frequency counting for grouping anagrams?

Sorting each string takes O(k log k) time per string, where k is the length of the string. Frequency counting is more efficient with a time complexity of O(k) per string, making it faster for large inputs.

What are the space complexities of the sorting and frequency count methods?

Both methods use O(n) space for the hash table to store the grouped anagrams, where n is the number of strings. The sorting method may also use additional space for sorting the strings.

#49

Medium

auto_awesome数组·哈希·扫描

LeetCode 题解工作台

字母异位词分组

给你一个字符串数组，请你将字母异位词组合在一起。可以按任意顺序返回结果列表。示例 1: 输入: strs = ["eat", "tea", "tan", "ate", "nat", "bat"] 输出: [["bat"],["nat","tan"],["ate","eat","tea"]] 解…

数组哈希表字符串排序

题目描述

给你一个字符串数组，请你将字母异位词组合在一起。可以按任意顺序返回结果列表。

示例 1:

输入: strs = ["eat", "tea", "tan", "ate", "nat", "bat"]

输出: [["bat"],["nat","tan"],["ate","eat","tea"]]

解释：

在 strs 中没有字符串可以通过重新排列来形成 "bat"。
字符串 "nat" 和 "tan" 是字母异位词，因为它们可以重新排列以形成彼此。
字符串 "ate" ，"eat" 和 "tea" 是字母异位词，因为它们可以重新排列以形成彼此。

示例 2:

输入: strs = [""]

输出: [[""]]

示例 3:

输入: strs = ["a"]

输出: [["a"]]

提示：

1 <= strs.length <= 10⁴
0 <= strs[i].length <= 100
strs[i] 仅包含小写字母

lightbulb

解题思路

方法一：哈希表

遍历字符串，对每个字符串按照字符字典序排序，得到一个新的字符串。
以新字符串为 key，[str] 为 value，存入哈希表当中（HashMap<String, List<String>>）。
后续遍历得到相同 key 时，将其加入到对应的 value 当中即可。

以 strs = ["eat", "tea", "tan", "ate", "nat", "bat"] 为例，遍历结束时，哈希表的状况：

key	value
`"aet"`	`["eat", "tea", "ate"]`
`"ant"`	`["tan", "nat"]`
`"abt"`	`["bat"]`

最后返回哈希表的 value 列表即可。

时间复杂度 $O(n\times k\times \log k)$ 。其中 $n$ 和 $k$ 分别是字符串数组的长度和字符串的最大长度。

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class Solution:
    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
        d = defaultdict(list)
        for s in strs:
            k = ''.join(sorted(s))
            d[k].append(s)
        return list(d.values())

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Do you understand how hash tables can be used to group strings efficiently?
question_mark
Can you explain the trade-off between sorting strings and using frequency counts?
question_mark
Will you consider edge cases such as empty strings and strings with single characters?

warning

常见陷阱

外企场景

error
Forgetting to handle empty strings, which should be grouped correctly.
error
Not efficiently handling large inputs, especially when the number of strings is high (up to 10^4).
error
Assuming the problem only requires sorting the input, rather than considering the hashing approach.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Group Anagrams with case sensitivity
arrow_right_alt
Group Anagrams with longer strings up to length 100
arrow_right_alt
Group Anagrams with additional constraints like handling non-English characters

help

常见问题

外企场景

继续练习

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