time-complexity

问题 I am trying to solve an interesting problem, it's easy to just do a groupBy for aggregation like sum, count etc. But this problem is slightly different. Let me explain: This is my list of tuples: val repeatSmokers: List[(String, String, String, String, String, String)] = List( ("ID76182", "sachin", "kita MR.", "56308", "1990", "300"), ("ID76182", "KOUN", "Jana MR.", "56714", "1990", "100"), ("ID76182", "GANGS", "SKILL", "27539", "1990", "255"), ("ID76182", "GANGS", "SKILL", "27539", "1990",

问题 How can I calculate the time complexity of zip()? testList = [[1,2,3]for _ in range(5)] zip(*testList) 回答1: Assume you zip N iterables. In python 3.x , the zip function itself runs in O(1) time, as it just allocates a special iterable (called the zip object), and assigns the parameter array to an internal field. The function invocation itself (before control reaches in zip) is O(N) , as the interpreter must convert the parameters to an array. Every subsequent next call on the iterator also

问题 How can I calculate the time complexity of zip()? testList = [[1,2,3]for _ in range(5)] zip(*testList) 回答1: Assume you zip N iterables. In python 3.x , the zip function itself runs in O(1) time, as it just allocates a special iterable (called the zip object), and assigns the parameter array to an internal field. The function invocation itself (before control reaches in zip) is O(N) , as the interpreter must convert the parameters to an array. Every subsequent next call on the iterator also

问题 Question 1 for (i = 0; i < n; i++) { for (j = 0; j < i * i ; j++){ } } Answer: O(n^3) At first glance, O(n^3) made sense to me, but I remember a previous problem I did: Question 2 for (int i = n; i > 0; i /= 2) { for (int j = 0; j < i; j++) { //statement } } Answer: O(n) For Question 2, the outer loop is O(log n) and the inner loop is O(2n / log n) resulting in O(n). The inner loop is O(2n / log n) because - see explanation here: Big O of Nested Loop (int j = 0; j < i; j++) Why we don't do

问题 Given a string, I know how to find the number of palindromic substrings in linear time using Manacher's algorithm. But now I need to find the number of distinct/unique palindromic substrings. Now, this might lead to an O(n + n^2) algorithm - one 'n' for finding all such substrings, and n^2 for comparing each of these substrings with the ones already found, to check if it is unique. I am sure there is an algorithm with better complexity. I was thinking of maybe trying my luck with suffix trees

问题 I was trying to solve the 3 sum problem in cpp. Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero. class Solution { public: vector<vector<int>> threeSum(vector<int>& nums) { int size = nums.size(); vector<vector<int>> result; for (int i = 0; i < size - 2; ++i) { for (int j = i + 1; j < size - 1; ++j) { for (int k = j + 1; k < size; ++k) { if (sumToZero(i, j, k, nums)) { vector<int> newComb

问题 How can I find the complexity of a Ruby method? For example length? If I look at the source code, I see this: static VALUE rb_ary_length(VALUE ary) { long len = RARRAY_LEN(ary); return LONG2NUM(len); } But I don't know how to read that in order to find the Big O notation. 回答1: There is no maintained list of theoretical complexities of Ruby methods. Of interest to you is minitest/benchmark , which is used like this: require 'minitest/autorun' require 'minitest/benchmark' class TestFoobar <

问题 How can I find the complexity of a Ruby method? For example length? If I look at the source code, I see this: static VALUE rb_ary_length(VALUE ary) { long len = RARRAY_LEN(ary); return LONG2NUM(len); } But I don't know how to read that in order to find the Big O notation. 回答1: There is no maintained list of theoretical complexities of Ruby methods. Of interest to you is minitest/benchmark , which is used like this: require 'minitest/autorun' require 'minitest/benchmark' class TestFoobar <

问题 How can I find the complexity of a Ruby method? For example length? If I look at the source code, I see this: static VALUE rb_ary_length(VALUE ary) { long len = RARRAY_LEN(ary); return LONG2NUM(len); } But I don't know how to read that in order to find the Big O notation. 回答1: There is no maintained list of theoretical complexities of Ruby methods. Of interest to you is minitest/benchmark , which is used like this: require 'minitest/autorun' require 'minitest/benchmark' class TestFoobar <

问题 Function(int n) if(n<=2) return 1; for(i=n ; i>n/8 ; i-=n/2) for(j=n ; j>2 ; j=j/2) syso(); return Function(n/2); In order to calculate I have done the following : T(n) = T(n/2) + O(1) + 2logn T(n/2): the recursive call to the function. O(1) : the if statement. 2logn: the first "for" will run only 2 times * (the second "for") that will run logn times. ** I have assumed that the second for loop will divide the j by 2 meaning that I will have j/2^k times iterations = logn. With the same logic