big-o

问题 C program that finds starting and ending indexes of subarray to remove to make given array have equal sums of left and right side. If its impossible print -1. I need it to work for any array, this is just for testing. Heres a code that finds equilibrium. #include <stdio.h> int equilibrium(int arr[], int n) { int i, j; int leftsum, rightsum; /* Check for indexes one by one until an equilibrium index is found */ for (i = 0; i < n; ++i) { /* get left sum */ leftsum = 0; for (j = 0; j < i; j++)

问题 Closed . This question needs to be more focused. It is not currently accepting answers. Want to improve this question? Update the question so it focuses on one problem only by editing this post. Closed 6 years ago . Improve this question I am trying to understand what exactly Big O notation is. I understand it in the literal and practical sense by going through the answers of similar questions in SO. But what the answers don't explain and i fail to understand is the mathematical basis behind

问题 "Every comparison-based algorithm to sort n elements must take Ω(nlogn) comparisons in the worst case. With this fact, what would be the complexity of constructing a n-node binary search tree and why?" Based on this question, I am thinking that the construction complexity must be at least O(nlogn). That said, I can't seem to figure out how to find the total complexity of construction. 回答1: The title of the question and the text you quote are asking different things. I am going to address what

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 Does anyone know of any good resources to learn big o notation? In particular learning how to walk through some code and being able to see that it would be O(N^2) or O(logN)? Preferably something that can tell me why a code like this is equal to O(N log N) def complex(numbers): N = len(numbers) result = 0 for i in range(N): j = 1 while j < N: result += numbers[i]*numbers[j] j = j*2 return result Thanks! 回答1: To start, let me define to you what O(N log N) is. It means, that the program will

问题 When reasoning about runtime cost in a garbage collected language, what is the cost of a statement such as myList = null; in terms of 'n' (the number of elements in the list)? For sake of argument, consider the list to be a singly linked list of reference types with no finalisation required. More generally, I'm looking for any information on how runtime cost can be analysed in a language with GC. 回答1: My own thought is that the cost is likely to be either O(1) or O(n) depending on the

问题 I was reading about big O notation in java programming. I found the following table it shows different big O for different data structures. http://bigocheatsheet.com/ My questions are: If I want to delete an item in an array, is it O(n^2) ? (search and delete) If I want to delete an item in a stack, is it O(n) ? Which one is more effective, is it a single linked list or double single list? In what case is that the insert operation is O(1) or O(n) in a hash table? If I want to delete an item

问题 I was reading about big O notation in java programming. I found the following table it shows different big O for different data structures. http://bigocheatsheet.com/ My questions are: If I want to delete an item in an array, is it O(n^2) ? (search and delete) If I want to delete an item in a stack, is it O(n) ? Which one is more effective, is it a single linked list or double single list? In what case is that the insert operation is O(1) or O(n) in a hash table? If I want to delete an item

问题 I understand that O(N) is essentially equal to O(cN) where c='some constant'. But if N = c. Doesn't that make it O(N)^2. Does this hold as c increases, or is there some formal limit. 回答1: If N = c then c is not constant. Therefore this is never the case. 回答2: O(n) means that the runtime of the algorithm increases linearly with the size of the input. If you double the size of the input, you double the runtime. If you triple the size of the input, you triple the runtime. And so on. So the graph

问题 I have two integer arrays. I need to find out two numbers, one from each array, whose sum is equal to 2. This is very simple in O(n^2) but is there a way to do it faster? 回答1: You can do it in O(N+M) time and O(N) space like this: Put elements of array a into a hash set Walk through array b , and check if hash table contains 2-b[i] Constructing a hash set of N elements takes O(N) time and O(N) space. Checking each of M elements against the hash set takes O(1), for a total of O(N+M) time. 来源：