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For example, for the array of values [−2, 1, −3, 4, −1, 2, 1, −5, 4], the contiguous subarray with the largest sum is [4, −1, 2, 1], with sum 6. Some properties of this problem are: If the array contains all non-negative numbers, then the problem is trivial; a maximum subarray is the entire array.
Comparison of two revisions of an example file, based on their longest common subsequence (black) A longest common subsequence (LCS) is the longest subsequence common to all sequences in a set of sequences (often just two sequences).
The longest increasing subsequence problem is closely related to the longest common subsequence problem, which has a quadratic time dynamic programming solution: the longest increasing subsequence of a sequence is the longest common subsequence of and , where is the result of sorting.
Consider finding a shortest path for traveling between two cities by car, as illustrated in Figure 1. Such an example is likely to exhibit optimal substructure. That is, if the shortest route from Seattle to Los Angeles passes through Portland and then Sacramento, then the shortest route from Portland to Los Angeles must pass through Sacramento too.
The register width of a processor determines the range of values that can be represented in its registers. Though the vast majority of computers can perform multiple-precision arithmetic on operands in memory, allowing numbers to be arbitrarily long and overflow to be avoided, the register width limits the sizes of numbers that can be operated on (e.g., added or subtracted) using a single ...
List comprehension is a syntactic construct available in some programming languages for creating a list based on existing lists. It follows the form of the mathematical set-builder notation ( set comprehension ) as distinct from the use of map and filter functions.
Linear probing is a component of open addressing schemes for using a hash table to solve the dictionary problem.In the dictionary problem, a data structure should maintain a collection of key–value pairs subject to operations that insert or delete pairs from the collection or that search for the value associated with a given key.
The maximum sum is 1, attained by giving one agent the item with value 1 and the other agent nothing. But the max-min allocation gives each agent value at least e, so the sum must be at most 3e. Therefore the POF is 1/(3e), which is unbounded. Alice has two items with values 1 and e, for some small e>0. George has two items with value e. The ...