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Group strings that are anagrams

J.P. MorganSoftware Engineer · Onsite
EasySoftware EngineerOnsiteCoding & Algorithms

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This problem tests your ability to efficiently group strings based on their character counts and your knowledge of hash maps to solve anagram-related categorization problems.

Problem Statement

Given an array of strings, write a function to group all strings that are anagrams of each other together. An anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.

Function Signature

def group_anagrams(strs: list[str]) -> list[list[str]]:

Constraints

  • 0 <= len(strs) <= 10^4
  • 0 <= len(strs[i]) <= 100
  • strs[i] consists of lowercase English letters.

Examples

Example 1: Input: ["eat", "tea", "tan", "ate", "nat", "bat"] Output: [["bat"], ["nat", "tan"], ["ate", "eat", "tea"]] Explanation: The order of the groups or the internal list order does not matter.

Example 2: Input: [""] Output: [[""]]

Implementation Details

You should aim for an efficient time complexity, ideally better than O(n * k * log k) where n is the number of strings and k is the maximum length of a string, though both sorting and counting approaches are acceptable as starting points.

Hints

Hint 1

Consider that two strings are anagrams if and only if they result in the same sorted string.

Hint 2

Alternatively, you can represent each string by a frequency count of its characters (e.g., a tuple of 26 integers) to use as a hash map key.

Hint 3

Using a hash map allows you to group the strings in a single pass over the input array.

Follow-up questions

  • How would you modify your solution if the input contained Unicode characters instead of just lowercase English letters?
  • What is the space complexity of your solution, and can you optimize it if the number of anagram groups is small relative to the total number of strings?
  • If the strings were extremely long (e.g., length 10^6), would the frequency count approach be more efficient than sorting? Why?