space-complexity

问题 from typing import List def countTeams(num: int, skills: List[int], minAssociates: int, minLevel: int, maxLevel: int) -> int: # WRITE YOUR BRILLIANT CODE HERE qualified = [] possible_teams = [[]] for a in skills: if minLevel <= a <= maxLevel: qualified.append(a) num_teams = 0 while qualified: person = qualified.pop() new_teams = [] for team in possible_teams: print(team) print(person) new_team = [person] + team print(new_team) if len(new_team) >= minAssociates: num_teams += 1 new_teams.append

问题 I've the following code and I'm trying to get the time complexity. seen = set() a=[4,4,4,3,3,2,1,1,1,5,5] result = [] for item in a: if item not in seen: seen.add(item) result.append(item) print (result) As far as my understanding goes as I'm accessing the list the time complexity for that operation would be O(n) . As with the if block each time I've a lookup to the set and that would cost another O(n) . So is the overall time complexity O(n^2) ? Does the set.add() also add to the complexity?

问题 I am trying to understand Space Complexities of different sorting algorithms. http://bigocheatsheet.com/?goback=.gde_98713_member_241501229 from the above link I found that the complexity of bubble sort,insertion and selection sort is O(1) where as quick sort is O(log(n)) and merge sort is O(n). we were actually not allocating extra memory in any of the algorithms. Then why the space complexities are different when we are using the same array to sort them? 回答1: When you run code, memory is

问题 I am trying to understand Space Complexities of different sorting algorithms. http://bigocheatsheet.com/?goback=.gde_98713_member_241501229 from the above link I found that the complexity of bubble sort,insertion and selection sort is O(1) where as quick sort is O(log(n)) and merge sort is O(n). we were actually not allocating extra memory in any of the algorithms. Then why the space complexities are different when we are using the same array to sort them? 回答1: When you run code, memory is

问题 I am trying to understand Space Complexities of different sorting algorithms. http://bigocheatsheet.com/?goback=.gde_98713_member_241501229 from the above link I found that the complexity of bubble sort,insertion and selection sort is O(1) where as quick sort is O(log(n)) and merge sort is O(n). we were actually not allocating extra memory in any of the algorithms. Then why the space complexities are different when we are using the same array to sort them? 回答1: When you run code, memory is

问题 I have been trying to solve this problem in an efficient way. The problem is: Problem Statement Given a list of tuples in the form [(start1, end1), (start2, end2), (start3, end3)....(startn, endn)] where start and end are positive integers. Each tuple is to represent a time window, for example: [(1, 3), (73, 80)...] . Find the time (integer) where max concurrency occurs and get the tuples where max concurrency occurs. Constraints: start and end are integers of time and are between 0 to n For

问题 I have been trying to solve this problem in an efficient way. The problem is: Problem Statement Given a list of tuples in the form [(start1, end1), (start2, end2), (start3, end3)....(startn, endn)] where start and end are positive integers. Each tuple is to represent a time window, for example: [(1, 3), (73, 80)...] . Find the time (integer) where max concurrency occurs and get the tuples where max concurrency occurs. Constraints: start and end are integers of time and are between 0 to n For

问题 I'm working on the problem stated in the question statement. I know my solution is correct (ran the program) but I'm curious as to whether or not I'm analyzing my code (below) correctly. def parens(num) return ["()"] if num == 1 paren_arr = [] parens(num-1).each do |paren| paren_arr << paren + "()" unless "()#{paren}" == "#{paren}()" paren_arr << "()#{paren}" paren_arr << "(#{paren})" end paren_arr end parens(3), as an example, will output the following: ["()()()", "(()())", "(())()", "()(())

问题 Consider the following bubble sort program: arr = map(int, raw_input().split(' ')) print "Unsorted: \n{arr_name}".format(arr_name = arr) for j in range(len(arr) - 1, 0, -1): for i in range(j): if (arr[i] > arr[i + 1]): arr[i], arr[i + 1] = arr[i +1], arr[i] print "Sorted: \n{arr_name}".format(arr_name = arr) Usually a temp variable is used for sorting and that would be mean a space complexity of 0(1) . But my understanding is that, tuple swap is only a reassignment of identifiers to objects

问题 I am reading about the KMP substring search algorithm and the examples I find online use an one-dimensional table to build the prefix information table. I also read the Sedgewick explanation and he used a 2-D array to build the table and explicitly states that the space complexity of KMP is O(RM) where R is the alphabet size and M the pattern size while everywhere else it is stated that the space complexity is just O(M + N) i.e. the text to process and the pattern size itself. So I am