big-o

问题 For example: 'hello'.count('e') Is this O(n)? I'm guessing the way it works is it scans 'hello' and increments a counter each time the letter 'e' is seen. How can I know this without guessing? I tried reading the source code here, but got stuck upon finding this: def count(s, *args): """count(s, sub[, start[,end]]) -> int Return the number of occurrences of substring sub in string s[start:end]. Optional arguments start and end are interpreted as in slice notation. """ return s.count(*args)

问题 For example: 'hello'.count('e') Is this O(n)? I'm guessing the way it works is it scans 'hello' and increments a counter each time the letter 'e' is seen. How can I know this without guessing? I tried reading the source code here, but got stuck upon finding this: def count(s, *args): """count(s, sub[, start[,end]]) -> int Return the number of occurrences of substring sub in string s[start:end]. Optional arguments start and end are interpreted as in slice notation. """ return s.count(*args)

问题 Classic problem with 12 coins ( or marbles) one of which is fake. Fake coin assumed to be lighter than real one. Having scales to compare coins (or marbles). One can do comparison one by one and compare all 12 coins. More efficiently one can do it using Decrease By Factor algorithm. Which is divide the coin stack by 2 and compare 2 stacks on the scales. Big O of the decrease by factor 2 is log2n. There is more efficient decrease by factor 3 (log3n) algorithm but I have not yet found it. If

问题 Classic problem with 12 coins ( or marbles) one of which is fake. Fake coin assumed to be lighter than real one. Having scales to compare coins (or marbles). One can do comparison one by one and compare all 12 coins. More efficiently one can do it using Decrease By Factor algorithm. Which is divide the coin stack by 2 and compare 2 stacks on the scales. Big O of the decrease by factor 2 is log2n. There is more efficient decrease by factor 3 (log3n) algorithm but I have not yet found it. If

问题 Suppose that N and M are two parameters of an algorithm. Is the following simplification correct? O(N+NM) = O[N(1+M)] = O(NM) In other words, is it allowed to remove the constant in such a context? 回答1: TL;DR Yes Explanation By the definition of the Big-Oh notation, if a term inside the O(.) is provably smaller than a constant times another term for all sufficiently large values of the variable, then you can drop the smaller term. You can find a more precise definition of Big-Oh here, but

问题 Suppose that N and M are two parameters of an algorithm. Is the following simplification correct? O(N+NM) = O[N(1+M)] = O(NM) In other words, is it allowed to remove the constant in such a context? 回答1: TL;DR Yes Explanation By the definition of the Big-Oh notation, if a term inside the O(.) is provably smaller than a constant times another term for all sufficiently large values of the variable, then you can drop the smaller term. You can find a more precise definition of Big-Oh here, but

问题 currently I have best case as o(n), worst case as o(n). I am unsure if this is correct for the following code segment and I would really appreciate if someone could confirm it is correct, or explain why it's wrong. Thanks! template<typename T> void vector_print(const T& vector, bool repeat) { for (typename T::size_type i = 0; i < vector.size(); ++i) { for (typename T::size_type j = 0; j < vector.size(); ++j) { std::cout << vector[j] << ","; } std::cout << std::endl; if (!repeat) { break; }

问题 From Wikipedia: O(|E| + |V| log|V|) From Big O Cheat List: O((|V| + |E|) log |V|) I consider there is a difference between E + V log V and (E+V) log V , isn't there? Because, if Wikipedia's one is correct, shouldn't it be shown as O(|V| log |V|) only then (Removing |E| ) for a reason I do not understand?)? What is the Big O of Dijkstra with Fibonacci-Heap? 回答1: The complexity of Dijkstra's shortest path algorithm is: O(|E| |decrease-key(Q)| + |V| |extract-min(Q)|) where Q is the min-priority

问题 From Wikipedia: O(|E| + |V| log|V|) From Big O Cheat List: O((|V| + |E|) log |V|) I consider there is a difference between E + V log V and (E+V) log V , isn't there? Because, if Wikipedia's one is correct, shouldn't it be shown as O(|V| log |V|) only then (Removing |E| ) for a reason I do not understand?)? What is the Big O of Dijkstra with Fibonacci-Heap? 回答1: The complexity of Dijkstra's shortest path algorithm is: O(|E| |decrease-key(Q)| + |V| |extract-min(Q)|) where Q is the min-priority

问题 From Wikipedia: O(|E| + |V| log|V|) From Big O Cheat List: O((|V| + |E|) log |V|) I consider there is a difference between E + V log V and (E+V) log V , isn't there? Because, if Wikipedia's one is correct, shouldn't it be shown as O(|V| log |V|) only then (Removing |E| ) for a reason I do not understand?)? What is the Big O of Dijkstra with Fibonacci-Heap? 回答1: The complexity of Dijkstra's shortest path algorithm is: O(|E| |decrease-key(Q)| + |V| |extract-min(Q)|) where Q is the min-priority