Bit-Manipulation

Common bit operations, AND only removes bits while OR only adds them, finding adjacent ones with n & (n » 1), and all-ones numbers 2^k - 1

Building a Trie from nested dicts, counting on nodes, pairing it with DFS, and when a Trie is not actually needed

LeetCode 864 note: the BFS state is (position, bitmask of keys), so the same cell with different keys is a different node

LeetCode 1178 note: a word only matters by its set of letters, so count 26-bit masks and enumerate each puzzle’s submasks

LeetCode 3950 note: each 1 in n & (n » 1) marks a pair of adjacent ones, so exactly one pair means it has a single 1

LeetCode 3910 note: n is tiny, so enumerate subsets with a bitmask, check the even sum first, then check that the induced subgraph is connected

LeetCode 3827 note: the only numbers with all-equal bits are 0 and 2^k - 1, so the count is (n + 1).bit_length()

LeetCode 2401 note: pairwise AND 0 ⇔ each bit is used by at most one number; a mask holds the window’s bits, and on a conflict the left shrinks and its bits are removed

LeetCode 1763 note: a character whose other case never appears cannot be in the answer, so split on it and recurse, preferring the left side on ties

LeetCode 1461 note: put every length-k substring in a set and check that its size reaches 2^k; the last start is len(s) - k

LeetCode 1016 note: s has at most 1000 characters, so there are few substrings of any fixed length and a large n must be False; checking from n downward fails fast, or extend up to 30 bits from each start to collect every value <= n

LeetCode 1486 note: generate start + 2i by definition and XOR them together