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LeetCode LLC, doing business as LeetCode, is an online platform for coding interview preparation. The platform provides coding and algorithmic problems intended for users to practice coding . [ 1 ] LeetCode has gained popularity among job seekers in the software industry and coding enthusiasts as a resource for technical interviews and coding ...
LeetCode: LeetCode has over 2,300 questions covering many different programming concepts and offers weekly and bi-weekly contests. The programming tasks are offered in English and Chinese. Project Euler [18] Large collection of computational math problems (i.e. not directly related to programming but often requiring programming skills for ...
The Boy or Girl paradox surrounds a set of questions in probability theory, which are also known as The Two Child Problem, [1] Mr. Smith's Children [2] and the Mrs. Smith Problem. The initial formulation of the question dates back to at least 1959, when Martin Gardner featured it in his October 1959 "Mathematical Games column" in Scientific ...
The longest alternating subsequence problem has also been studied in the setting of online algorithms, in which the elements of are presented in an online fashion, and a decision maker needs to decide whether to include or exclude each element at the time it is first presented, without any knowledge of the elements that will be presented in the future, and without the possibility of recalling ...
An "eleet hacker" (31337 H4XØR) laptop sticker, along with a "Free Kevin [Mitnick]" sticker. Leet (or "1337"), also known as eleet or leetspeak, or simply hacker speech, is a system of modified spellings used primarily on the Internet.
def ternary_search (f, left, right, absolute_precision)-> float: """Find maximum of unimodal function f() within [left, right]. To find the minimum, reverse the if/else statement or reverse the comparison. """ while abs (right-left) >= absolute_precision: left_third = left + (right-left) / 3 right_third = right-(right-left) / 3 if f (left_third) < f (right_third): left = left_third else: right ...