Group Anagrams

Problem Statement

You’re a word organizer with a list of strings. Your task: group all anagrams together. This medium-level hashing challenge is a linguistic sorting spree—use a map to cluster those scrambled siblings!

Example

Input: strs = ["eat", "tea", "tan", "ate", "nat", "bat"]

Output: [["eat", "tea", "ate"], ["tan", "nat"], ["bat"]]

Input: strs = [""]

Output: [[""]]

Input: strs = ["a"]

Output: [["a"]]

Code

public class Solution {
    public List> groupAnagrams(String[] strs) {
        Map> map = new HashMap<>();
        for (String str : strs) {
            char[] chars = str.toCharArray();
            Arrays.sort(chars);
            String key = new String(chars);
            map.computeIfAbsent(key, k -> new ArrayList<>()).add(str);
        }
        return new ArrayList<>(map.values());
    }
}
            

from collections import defaultdict
def group_anagrams(strs):
    anagrams = defaultdict(list)
    for s in strs:
        key = ''.join(sorted(s))
        anagrams[key].append(s)
    return list(anagrams.values())
            

function groupAnagrams(strs) {
    let map = new Map();
    for (let str of strs) {
        let key = str.split('').sort().join('');
        if (!map.has(key)) map.set(key, []);
        map.get(key).push(str);
    }
    return Array.from(map.values());
}
            

Explanation

• Hashing Insight: Sorted string as key groups anagrams together.
• Flow: Sort each string’s chars to create a unique key, map it to a list of originals.
• Example Walkthrough: "eat" → "aet", maps to ["eat", "tea", "ate"].
• Python Tip: defaultdict simplifies initialization.
• Alternative: Use char count arrays as keys for O(nk) without sorting.

Note