47. Group Anagrams

mediumAsked at Salesforce

Group strings that are anagrams of each other. Salesforce uses this to test canonical-form hashing — they want O(n*k) with a frequency-tuple key.

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

Source citations

Public interview reports confirming this problem appears in Salesforce loops.

Glassdoor (2026-Q1)— Salesforce uses anagram-like canonicalization in their deduplication-by-content logic.
Blind (2025-12)— Common Salesforce onsite question.

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. Sort each string as key

Sort the string's characters; use the sorted version as a hash key.

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

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

Tradeoff: Sort per string costs O(k log k). Acceptable but Salesforce prefers the frequency version.

2. Frequency tuple as key

For each string, build a 26-int count array; serialize as key.

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

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

Tradeoff: O(k) per string vs O(k log k) for sort. The frequency vector is the canonical form.

Salesforce-specific tips

Salesforce uses canonical-form hashing in their record-deduplication logic — same data, different field-order, must hash identically. They grade on whether you reach for the count-based canonicalization without prompting. Bonus signal: discuss that for very long strings, the frequency-vector approach wins; for short strings, sorting can be faster due to cache effects.

Common mistakes

Using the string itself as key — only catches identical strings, not anagrams.
Joining the count array without separator — '1', '11' collide for ['a'] and ['a','a'].
Forgetting to handle empty strings — works but worth confirming.

Follow-up questions

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

Valid Anagram (LC 242) — pairwise check.
Find All Anagrams in a String (LC 438).
Generalize to Unicode strings — Map<string, count>.

Solve it now

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Output

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FAQ

Why include separator in the key?

Because count.join('') makes '1' and '11' collide. Using ',' as separator gives unique keys per count tuple.

Sort key or frequency key — which is better?

Frequency key is O(k) vs sort's O(k log k). For long strings (k > 50), frequency wins. For short strings, sort can be faster due to JS engine optimizations.

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