Reviewed by Aditya Kumar · Last reviewed 2026-03-24
**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...
Pro-Move: Unicode normalization for i18n. Red Flag: O(n²) pairwise compare.
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. Production: Handle case, unicode.
Pro-Move: Unicode normalization for i18n. Red Flag: O(n²) pairwise compare.
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