12. Group Anagrams
mediumAsked at PostmanGroup strings into lists where each list contains anagrams of each other.
By Alex Chen, Founder, InterviewChamp.AI · Last verified
Problem
Given an array of strings, group the anagrams together. Return the answer in any order; the groups themselves can be in any order.
Constraints
1 <= strs.length <= 10^40 <= strs[i].length <= 100strs[i] consists of lowercase English letters
Examples
Example 1
strs = ["eat","tea","tan","ate","nat","bat"][["bat"],["nat","tan"],["ate","eat","tea"]]Example 2
strs = [""][[""]]Approaches
1. Sort key
Use the sorted string as a hash map key; group originals under that key.
- Time
- O(n * k log k)
- Space
- O(n*k)
const map = new Map();
for (const s of strs) {
const k = [...s].sort().join('');
if (!map.has(k)) map.set(k, []);
map.get(k).push(s);
}
return [...map.values()];Tradeoff:
2. Frequency-tuple key
Bucket each string by its 26-element char count serialized as the key, avoiding the sort.
- Time
- O(n * k)
- Space
- O(n*k)
function group(strs) {
const map = new Map();
for (const s of strs) {
const c = new Array(26).fill(0);
for (const ch of s) c[ch.charCodeAt(0) - 97]++;
const k = c.join(',');
if (!map.has(k)) map.set(k, []);
map.get(k).push(s);
}
return [...map.values()];
}Tradeoff:
Postman-specific tips
Postman engineers think in terms of canonical-key bucketing because grouping equivalent requests (same path + sorted query keys) is exactly how the proxy collapses duplicates.
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