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Maximum subarray problems arise in many fields, such as genomic sequence analysis and computer vision.. Genomic sequence analysis employs maximum subarray algorithms to identify important biological segments of protein sequences that have unusual properties, by assigning scores to points within the sequence that are positive when a motif to be recognized is present, and negative when it is not ...
Manacher (1975) invented an ()-time algorithm for listing all the palindromes that appear at the start of a given string of length . However, as observed e.g., by Apostolico, Breslauer & Galil (1995) , the same algorithm can also be used to find all maximal palindromic substrings anywhere within the input string, again in O ( n ) {\displaystyle ...
This algorithm runs in () time. The array L stores the length of the longest common suffix of the prefixes S[1..i] and T[1..j] which end at position i and j, respectively. The variable z is used to hold the length of the longest common substring found so far. The set ret is used to hold the set of strings which are of length z.
Chowdhury and Ramachandran devised a quadratic-time linear-space algorithm [9] [10] for finding the LCS length along with an optimal sequence which runs faster than Hirschberg's algorithm in practice due to its superior cache performance. [9] The algorithm has an asymptotically optimal cache complexity under the Ideal cache model. [11]
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 string spelled by the edges from the root to such a node is a longest repeated substring. The problem of finding the longest substring with at least k {\displaystyle k} occurrences can be solved by first preprocessing the tree to count the number of leaf descendants for each internal node, and then finding the deepest node with at least k ...
This leads to a size of log n / 4 for every tree. To enable random access in constant time to any tree, the trees not contained in the original array need to be included as well. An array with indices of log n / 4 bits length has size 2 log n / 4 = O (n). Example of Cartesian trees for A = [0,5,2,5,4,3,1,6,3]. Notice ...
For example, suppose Alice has two items with values 1 and e, for some small e>0. George has two items with value e. The capacity is 1. The maximum sum is 1 - when Alice gets the item with value 1 and George gets nothing. But the max-min allocation gives both agents value e. Therefore the POF is 1/(2e), which is unbounded. In both cases, if the ...