18. Group Anagrams

mediumAsked at Indeed

Cluster strings that are anagrams of each other — directly models how Indeed groups near-duplicate job titles before deduplication and canonical-name resolution.

By Alex Chen, Founder, InterviewChamp.AI · Last verified 2026-05-20

Problem

Given an array of strings strs, group the anagrams together. Return the groups in any order; each group may also be 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

[[""]]

Approaches

1. Sort each string as key

Sorted characters form a canonical key; group all strings sharing the same sorted key.

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. Character frequency key

Encode each string as a 26-integer frequency array to avoid the sort cost, dropping per-string time from O(k log k) to O(k).

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

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

Tradeoff:

Indeed-specific tips

Indeed interviewers reward the frequency-key variant because it mirrors how their job-title normalizer avoids sorts on high-volume streams — explain the trade-off explicitly.

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Output

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