Group Anagrams: the idea to remember.

The core insight

Use a frequency signature as the hash map key.

Why it works

Anagrams have the same count in every character bucket, so they map to exactly the same key.

The mistake to avoid

Concatenating counts without separators can create ambiguous signatures.

Commented Java solution

Use java.util.* imports and a surrounding solution class. Tree snippets assume the standard TreeNode definition; linked lists use ListNode and Clone Graph uses a Node with val and neighbors; Copy Random List uses Node with val, next and random. Read the input assumptions above.

List<List<String>> groupAnagrams(String[] words) {
    Map<String, List<String>> groups = new HashMap<>();
    for (String word : words) {
        int[] count = new int[26];
        for (char c : word.toCharArray()) count[c - 'a']++;
        // Delimited serialization distinguishes [1,11] from [11,1].
        String key = Arrays.toString(count);
        groups.computeIfAbsent(key, k -> new ArrayList<>()).add(word);
    }
    return new ArrayList<>(groups.values());
}

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