space-complexity

问题 Algorithm reorder array according to given index a[] = [50, 40, 70, 60, 90] index[] = [3, 0, 4, 1, 2] a= [60,50,90,40,70] in O(n) and With out extra array/spaces 回答1: You'll need space for a temp variable and loop counters / indices. The usual "reorder" according to algorithm is also going to change index[] back to {0, 1, 2, 3, 4}. Hint, noting the ordering of indices in index[]. {0, 1, 2, 3, 4} index[] = {3, 0, 4, 1, 2} The reordering can be done by following the "cycles". Start with index[0

问题 Algorithm reorder array according to given index a[] = [50, 40, 70, 60, 90] index[] = [3, 0, 4, 1, 2] a= [60,50,90,40,70] in O(n) and With out extra array/spaces 回答1: You'll need space for a temp variable and loop counters / indices. The usual "reorder" according to algorithm is also going to change index[] back to {0, 1, 2, 3, 4}. Hint, noting the ordering of indices in index[]. {0, 1, 2, 3, 4} index[] = {3, 0, 4, 1, 2} The reordering can be done by following the "cycles". Start with index[0

问题 Let's take this implementation of Merge Sort as an example void mergesort(Item a[], int l, int r) { if (r <= l) return; int m = (r+l)/2; mergesort(a, l, m); ------------ (1) mergesort(a, m+1, r); ------------(2) merge(a, l, m, r); a) The time complexity of this Merge Sort is O(nlg(n)). Will parallelizing (1) and (2) give any practical gain ? Theorotically, it appears that after parallelizing them also you would end up in O(nlg(n). But practically can we get any gains ? b) Space complexity of

问题 I think defenitions of time complexity and space complexity for Turing machines are identical and I can't differentiate between them. Please help me. Thanks. 回答1: With regards to a Turing machine, time complexity is a measure of how many times the tape moves when the machine is started on some input. Space complexity refers to how many cells of the tape are written to when the machine runs. The time complexity of a TM is connected to its space complexity. In particular, if tue space

问题 What space complexity does the python sort take? I can't find any definitive documentation on this anywhere 回答1: Space complexity is defined as how much additional space the algorithm needs in terms of the N elements. And even though according to the docs, the sort method sorts a list in place, it does use some additional space, as stated in the description of the implementation: timsort can require a temp array containing as many as N//2 pointers, which means as many as 2*N extra bytes on 32

问题 Mathematica' CylindricalDecomposition implements an algorithm known as Cylindrical Algebraic Decomposition. Wolfram MathWorld's article on Cylindrical Algebraic Decomposition says that this algorithm "becomes computationally infeasible for complicated inequalities." Can this statement be made more precise? Specifically, how does the time and space relate to the degree and number of variables of the multivariate polynomials? Does the time and space depend on other parameters? 回答1: Tarski

问题 Using Bloom filter, we will be getting space optimization. The cassandra framework also has an implementation of Bloom Filter. But in detail, how is this space optimization achieved? 回答1: A bloom filter isn't a "framework". It's really more like simply an algorithm. The implementation ain't very long. Here's one in Java I've tried ( .jar , source code and JavaDoc being all available): "Stand alone Java implementations of Cuckoo Hashing and Bloom Filters" (you may want to Google for this in

问题 Given an array where number of occurrences of each number is odd except one number whose number of occurrences is even. Find the number with even occurrences. e.g. 1, 1, 2, 3, 1, 2, 5, 3, 3 Output should be: 2 The below are the constraints: Numbers are not in range. Do it in-place. Required time complexity is O(N). Array may contain negative numbers. Array is not sorted. With the above constraints, all my thoughts failed: comparison based sorting, counting sort, BST's, hashing, brute-force. I

问题 Given an array where number of occurrences of each number is odd except one number whose number of occurrences is even. Find the number with even occurrences. e.g. 1, 1, 2, 3, 1, 2, 5, 3, 3 Output should be: 2 The below are the constraints: Numbers are not in range. Do it in-place. Required time complexity is O(N). Array may contain negative numbers. Array is not sorted. With the above constraints, all my thoughts failed: comparison based sorting, counting sort, BST's, hashing, brute-force. I

问题 I am curious to know which algorithm is better : Algorithm with O(n log n) time and O(1) space complexity Algorithm with O(n) time and O(n) space complexity Most of the algorithm which are solved in O(n long n) time and constant space can be solved in O(n) time by paying penalty in terms of space. Which algorithm is better ? How do I decide between these two parameters ? Example : Array Pair Sum Can be solved in O(n logn) time by sorting Can be solved using hash maps in O(n) time but with O(n