Bit-Manipulation

LeetCode 137 note: with everything else appearing three times, count the 1s at each bit modulo 3; the leftover bits form the answer, minding negatives

LeetCode 287 note: treat i -> nums[i] as a linked list; the duplicate is the cycle entry, found with Floyd’s algorithm in O(1) space without modifying the array

LeetCode 90 note: sort first; for a number equal to the previous one, extend only the subsets created in the previous step

LeetCode 78 note: each element is either taken or not; iteratively, each new number copies every existing subset with the number appended

LeetCode 1018 note: each new bit turns the prefix into x = 2x + bit, so keep only x mod 5 and the number never grows

LeetCode 868 note: remember the last 1’s position and take the maximum distance at each new 1

LeetCode 832 note: reverse each row and invert its bits, which two pointers can do at once

LeetCode 762 note: numbers are below 10^6 < 2^20, so the set-bit count is at most 19; check it against a small set of primes

LeetCode 693 note: alternating bits means the binary string has neither 00 nor 11; in bits, n ^ (n » 1) is all ones

LeetCode 645 note: use sums: missing = sum of 1..n - sum of the set, duplicate = array sum - sum of the set

LeetCode 476 note: build an all-ones mask as long as num; num ^ mask is the complement

LeetCode 461 note: each 1 in x ^ y is a differing bit, so count the ones