Lesson 18 — Anagram check

Anagrams: compare letter counts

Two words are anagrams if they use exactly the same letters the same number of times ("listen" / "silent"). Instead of sorting, count each letter in a dict and compare the two dicts. d.get(key, 0) returns the stored value, or 0 if the key is new.

counts = {}
counts["a"] = counts.get("a", 0) + 1   # {'a': 1}

Counting is O(n); sorting is O(n log n).

Why it matters: "count with a dict, then compare" solves a whole family of problems: duplicates, most common item, matching inventories.