49. Group Anagrams

mediumAsked at Duolingo

Cluster strings that are anagrams of each other into groups — the hash-keying technique Duolingo's vocabulary engine uses to bundle conjugation variants of the same root word into a single learning card set.

By Alex Chen, Founder, InterviewChamp.AI · Last verified 2026-05-26

Problem

Given an array of strings strs, group the anagrams together. You can return the answer in any order. Two strings are anagrams if one can be formed by rearranging all characters of the other.

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"]]

Explanation: All anagrams share the same sorted form as their canonical key.

Example 2

Input

strs = [""]

Output

[[""]]

Example 3

Input

strs = ["a"]

Output

[["a"]]

Approaches

1. Sort-based grouping

For each string, sort its characters to form a canonical key and group strings by that key in a Map.

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

function groupAnagrams(strs) {
  const map = new Map();
  for (const s of strs) {
    const key = s.split('').sort().join('');
    if (!map.has(key)) map.set(key, []);
    map.get(key).push(s);
  }
  return [...map.values()];
}

Tradeoff:

2. Optimal — character-count key

Encode each string as a 26-integer frequency tuple; use that string as the Map key, eliminating the sort step.

Time: O(n * k)
Space: O(n * k)

function groupAnagrams(strs) {
  const map = new Map();
  const a = 'a'.charCodeAt(0);
  for (const s of strs) {
    const count = new Array(26).fill(0);
    for (const ch of s) count[ch.charCodeAt(0) - a]++;
    const key = count.join(',');
    if (!map.has(key)) map.set(key, []);
    map.get(key).push(s);
  }
  return [...map.values()];
}

Tradeoff:

Duolingo-specific tips

Duolingo's word-grouping systems constantly cluster conjugation forms — 'run', 'ran', 'running' all map to a root. Interviewers expect you to know both the sort key and the frequency-count key approaches and to articulate when each wins: sort is simpler and O(k log k); count key is O(k) but produces a longer key string. A common follow-up is 'How would you handle Unicode characters or multi-character grapheme clusters?' — answer: use a Map with custom hash or normalize to code points before counting.

Solve it now

Free. No sign-up. Python and JavaScript run instantly in your browser.

Output

Press Run or Cmd+Enter to execute

Practice these live with InterviewChamp.AI

Drill Group Anagrams and other Duolingo interview questions under real-loop conditions with instant feedback on your reasoning, complexity claims, and code.

Practice these live with InterviewChamp.AI →