Group Anagrams (LC 49) — Hash Map by Sorted Key

What does it solve?

Group a list of words so that anagrams land together: ["eat","tea","tan","ate","nat","bat"] → [["eat","tea","ate"],["tan","nat"],["bat"]]. The trick is finding a canonical key that's identical for all anagrams.

Key insight: two words are anagrams exactly when their sorted letters match. So sort each word to get a canonical key, then use a hash map from key → list of original words. Every word drops into its bucket in O(1) after an O(K log K) sort. This "group by canonical key" is the core grouping pattern.

The core idea

groups = {}
for word in words:
key = sorted(word) # canonical form
groups[key].append(word)
return groups.values()

Complexity

Time complexity

O(N·K log K)

N words, each of length up to K. Sorting each word is O(K log K); map insert is O(K) to hash the key.

Space complexity

O(N·K)

The map stores every word plus its key. An O(K) counting-key variant avoids the log factor.

Examples & edge cases

Live animation — sort each word into its bucket

current word sorted key new bucket existing bucket

Input words

Press Play to begin…

Hash map — sorted key → group of words

{ } empty

Current step detail

word = — key = — buckets = — groups = —

Step — of —

current word

—

being placed

sorted key

—

canonical form

buckets

—

distinct keys

words placed

—

of total

Select an example and press Play to start.

✓ Done

SpeedMedium

Step-by-step table

Step	word	sorted key	bucket	action	groups

C++ implementation

group_anagrams.cpp

#include <bits/stdc++.h>
using namespace std;

// Group words that are anagrams of each other (LeetCode 49)
vector<vector<string>> groupAnagrams(vector<string>& words) {
    unordered_map<string, vector<string>> groups;

    for (string& w : words) {
        string key = w;
        sort(key.begin(), key.end());   // canonical key: sorted letters
        groups[key].push_back(w);        // drop into its bucket
    }

    vector<vector<string>> res;
    for (auto& [key, grp] : groups) res.push_back(grp);
    return res;
}

int main() {
    vector<string> w = {"eat","tea","tan","ate","nat","bat"};
    for (auto& g : groupAnagrams(w)) {
        for (auto& s : g) cout << s << " ";
        cout << "\n";
    }
    return 0;
}

My Notes

📝Notes✅

▼

●Start typing to save...0 chars

Practice this problem