How do you approach Group Anagrams?

Two strings are anagrams if their sorted versions are equal.

What is the time complexity of Group Anagrams?

Time complexity: O(n k log k) where k is the max string length. Space complexity: O(n k).

What are common mistakes on Group Anagrams?

Storing the current value before checking for its complement, which lets an element pair with itself. Reaching for a sort to deduplicate when a set does it in one pass - that swap quietly costs you the O(n). Using a plain object as the map when keys may be numbers or __proto__; a real hash-map type keeps key identity intact.

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Group Anagrams

MediumData Structures

Description

Given an array of strings strs, group the anagrams together. You can return the answer in any order.

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.

Example 1

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

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

Example 2

Input: strs = [""]

Output: [[""]]

Example 3

Input: strs = ["a"]

Output: [["a"]]

Constraints

- 1 <= strs.length <= 10^4 - 0 <= strs[i].length <= 100 - strs[i] consists of lowercase English letters.

Group Anagrams: approach, complexity and pitfalls

A medium arrays & hashing problem, graded against 6 test cases (3 of them hidden).

The pattern: Arrays & Hashing

Hash maps and sets to trade memory for time - the most common first instinct in interviews.

Reach for it when you see: "Have I seen this before?", counting occurrences, or an O(n²) scan you want to make O(n).

More Arrays & Hashing problems →

If you are stuck

Show the 3-step hint ladder

Two strings are anagrams if their sorted versions are equal.
Use a hash map with the sorted string as the key.
Group all strings with the same sorted key together.

The optimal approach

Show the approach: Hash Map with Sorted Key

Sort each string to create a canonical form. Use this as a key in a hash map to group anagrams.

Time: O(n * k log k) where k is the max string length
Space: O(n * k)

The full reference solution in every supported language stays in the editor above - reveal it there once you have had a real attempt.

What the tests check

Read off this problem's own test suite, so these are the cases a submission actually has to survive.

Single element. A one-element case is included. Two-pointer and window solutions frequently return the wrong thing when the two indices start out equal.
Duplicate values. Repeated values appear in the input, so any logic that assumes distinct elements - or that skips over equal neighbours - is exercised.
Empty string. An empty string is passed, so index reads and any assumption of at least one character are tested.

Common Arrays & Hashing mistakes

These apply to the pattern as a whole, not just this problem.

Storing the current value before checking for its complement, which lets an element pair with itself.
Reaching for a sort to deduplicate when a set does it in one pass - that swap quietly costs you the O(n).
Using a plain object as the map when keys may be numbers or `__proto__`; a real hash-map type keeps key identity intact.