49. Group Anagrams

mediumAsked at GoDaddy

Cluster strings that are anagrams of each other — GoDaddy uses the same sorted-key hash-table pattern to group similar domain name variations and detect typo-squatting slugs in their marketplace.

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

Problem

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

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

Sort each string and use the sorted form as a key; compare every pair.

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

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

Tradeoff:

2. Character-count key

Build a 26-element frequency vector as the hash key; avoids sort and runs in O(k) per string.

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

function groupAnagrams(strs) {
  const map = new Map();
  for (const str of strs) {
    const count = new Array(26).fill(0);
    for (const ch of str) {
      count[ch.charCodeAt(0) - 97]++;
    }
    const key = count.join(',');
    if (!map.has(key)) map.set(key, []);
    map.get(key).push(str);
  }
  return Array.from(map.values());
}

Tradeoff:

GoDaddy-specific tips

GoDaddy values the character-count key upgrade over the naive sort because their domain-search engine processes millions of slug lookups per minute — shaving from O(k log k) to O(k) per string matters at SMB-hosting scale. State the tradeoff explicitly.

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Output

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