25. Group Anagrams

mediumAsked at Booking

Cluster strings that are rearrangements of each other — Booking uses the same canonical-key grouping technique when deduplicating hotel name variants and locale translations that share the same character 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.

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. Sort each word as key

Sort the characters of each string to get a canonical key; group by that key in a hash map.

Time: O(n * k log k)
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. Frequency vector as key

Build a 26-slot character count array for each string; stringify it as the hash key. Avoids the sort cost, O(n*k) total.

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

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

Tradeoff:

Booking-specific tips

Booking's internationalisation team asks follow-ups about Unicode strings — the sort-based approach works on any alphabet while the frequency-vector approach is hardcoded to 26 letters. Know when to use each.

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Output

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