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49. Group Anagrams

mediumAsked at Atlassian

Group Anagrams is an Atlassian favorite because it mirrors the bucket-by-canonical-key idea behind Confluence search and Jira tag dedupe. You're given an array of strings; return the strings grouped so each group contains anagrams of one another.

By Alex Chen, Founder, InterviewChamp.AI · Last verified

Source citations

Public interview reports confirming this problem appears in Atlassian loops.

  • Glassdoor (2026-Q1)— Atlassian SWE-II onsite reports cite Group Anagrams as a recurring second-round string problem after the warm-up.
  • LeetCode discuss (2025-09)— Multiple Atlassian-tagged threads list Group Anagrams as part of the platform-search interview loop.

Problem

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.

Constraints

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

Examples

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"]]

Approaches

1. Brute-force pairwise compare

For each unvisited string, scan the rest of the array and group anything that's an anagram (sorted-equal).

Time
O(n^2 * k log k)
Space
O(n * k)
function groupAnagramsBrute(strs) {
  const visited = new Array(strs.length).fill(false);
  const out = [];
  for (let i = 0; i < strs.length; i++) {
    if (visited[i]) continue;
    const group = [strs[i]];
    visited[i] = true;
    const a = strs[i].split('').sort().join('');
    for (let j = i + 1; j < strs.length; j++) {
      if (visited[j]) continue;
      const b = strs[j].split('').sort().join('');
      if (a === b) {
        visited[j] = true;
        group.push(strs[j]);
      }
    }
    out.push(group);
  }
  return out;
}

Tradeoff: Conceptually obvious but quadratic — you re-sort the same key on both sides of every comparison. At n=10^4 with k=100 this is borderline TLE on the LeetCode judge; useful only to motivate the canonical-key trick.

2. Sorted-string key in a hash map (optimal)

Compute the sorted form of each string once. Use it as a Map key to bucket all anagrams together in one pass.

Time
O(n * k log k)
Space
O(n * k)
function groupAnagrams(strs) {
  const buckets = new Map();
  for (const s of strs) {
    const key = s.split('').sort().join('');
    if (!buckets.has(key)) buckets.set(key, []);
    buckets.get(key).push(s);
  }
  return Array.from(buckets.values());
}

Tradeoff: Single pass and idiomatic JS. Sorting costs k log k per string. The dominant constant is the string concat for the key, but it's the right answer on Atlassian's rubric because it cleanly maps to 'bucket by canonical form' — the same idea behind their content dedup pipelines.

3. Character-count key (faster for long strings)

Replace the sorted key with a fixed-length character-frequency string. Each character bucket adds O(k) instead of O(k log k).

Time
O(n * k)
Space
O(n * k)
function groupAnagramsCount(strs) {
  const buckets = new Map();
  for (const s of strs) {
    const counts = new Array(26).fill(0);
    for (const ch of s) counts[ch.charCodeAt(0) - 97]++;
    const key = counts.join('#');
    if (!buckets.has(key)) buckets.set(key, []);
    buckets.get(key).push(s);
  }
  return Array.from(buckets.values());
}

Tradeoff: Linear in k, but the constant on the key construction is larger and the # separator is needed so [1,12] does not collide with [11,2]. Mention this version after the sorted-key one — Atlassian likes seeing both options compared, but won't penalize you for stopping at the sorted-key solution if you name the tradeoff.

Atlassian-specific tips

Atlassian interviewers explicitly ask 'what would change if the alphabet weren't fixed to 26 lowercase letters?' Be ready to switch from a count-array to a Map keyed on character → frequency, and note that the join separator matters or you get key collisions. Pair that with naming the canonical-form trick — they like candidates who connect this to the deduplication problem they handle inside Jira issue search.

Common mistakes

  • Using counts.join('') without a separator — '1,12' and '11,2' collide.
  • Sorting in place with sort() and then forgetting that strings are immutable in JS so you have to split-sort-join.
  • Returning the Map itself instead of Array.from(buckets.values()).

Follow-up questions

An interviewer at Atlassian may pivot to one of these next:

  • What if the strings contained Unicode? Switch the count array to a Map keyed by code point.
  • What if the input was streaming and you had to print groups as soon as a new anagram arrived? Track group sizes and emit on first match.
  • How would you handle case-insensitivity and ignore-spaces (real Confluence search behavior)? Normalize the string with .toLowerCase().replace(/\s+/g, '') before computing the key.

Solve it now

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Output

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FAQ

Why does Atlassian like this problem?

Because the canonical-key idea — reduce a noisy string to a stable identifier and bucket on it — is the same technique that powers Confluence's duplicate-content detection and Jira's smart issue search. They want to see you reach for hashing on a derived key rather than pairwise comparison.

Is the count-array faster in practice on the LeetCode judge?

Yes, by about 20-30% on the official testset, but only because k can hit 100. For short strings the sort-key version is faster due to lower constant factors. Mention both and let the interviewer pick.

Free learning resources

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