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If the array contains all non-negative numbers, then the problem is trivial; a maximum subarray is the entire array. If the array contains all non-positive numbers, then a solution is any subarray of size 1 containing the maximal value of the array (or the empty subarray, if it is permitted).
The inapproximability results apply to all extensions of the maximum coverage problem since they hold the maximum coverage problem as a special case. The Maximum Coverage Problem can be applied to road traffic situations; one such example is selecting which bus routes in a public transportation network should be installed with pothole detectors ...
The most common variant of bucket sort operates on a list of n numeric inputs between zero and some maximum value M and divides the value range into b buckets each of size M/b. If each bucket is sorted using insertion sort , the sort can be shown to run in expected linear time (where the average is taken over all possible inputs). [ 3 ]
Firstly, computing median of an odd list is faster and simpler; while one could use an even list, this requires taking the average of the two middle elements, which is slower than simply selecting the single exact middle element. Secondly, five is the smallest odd number such that median of medians works.
LeetCode: LeetCode has over 2,300 questions covering many different programming concepts and offers weekly and bi-weekly contests. The programming tasks are offered in English and Chinese. Project Euler [18] Large collection of computational math problems (i.e. not directly related to programming but often requiring programming skills for ...
import random # this function checks whether or not the array is sorted def is_sorted (random_array): for i in range (1, len (random_array)): if random_array [i] < random_array [i-1]: return False return True # this function repeatedly shuffles the elements of the array until they are sorted def bogo_sort (random_array): while not is_sorted (random_array): random. shuffle (random_array) return ...
In this variant of the problem, which allows for interesting applications in several contexts, it is possible to devise an optimal selection procedure that, given a random sample of size as input, will generate an increasing sequence with maximal expected length of size approximately . [11] The length of the increasing subsequence selected by ...
The odd–even sort algorithm correctly sorts this data in passes. (A pass here is defined to be a full sequence of odd–even, or even–odd comparisons. The passes occur in order pass 1: odd–even, pass 2: even–odd, etc.) Proof: This proof is based loosely on one by Thomas Worsch. [6]