**Logic:** Anagrams = same char counts. **Approach:** Group by sorted string or char-frequency tuple. **Python:** `groups = defaultdict(list); [groups[tuple(sorted(Counter(s).items()))].append(s) for s in words]`. O(n·k log k) for n words, k max length. **Alternative:** Sort each as key. **Why:** Deduplication, fuzzy matching....
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