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A string-searching algorithm, sometimes called string-matching algorithm, is an algorithm that searches a body of text for portions that match by pattern. A basic example of string searching is when the pattern and the searched text are arrays of elements of an alphabet ( finite set ) Σ.
A string is a substring (or factor) [1] of a string if there exists two strings and such that =.In particular, the empty string is a substring of every string. Example: The string = ana is equal to substrings (and subsequences) of = banana at two different offsets:
In computer science, a longest common substring of two or more strings is a longest string that is a substring of all of them. There may be more than one longest common substring. Applications include data deduplication and plagiarism detection.
One possible definition of the approximate string matching problem is the following: Given a pattern string =... and a text string = …, find a substring ′, = ′ … in T, which, of all substrings of T, has the smallest edit distance to the pattern P.
P denotes the string to be searched for, called the pattern. Its length is m. S[i] denotes the character at index i of string S, counting from 1. S[i..j] denotes the substring of string S starting at index i and ending at j, inclusive. A prefix of S is a substring S[1..i] for some i in range [1, l], where l is the length of S.
string.find(string, substring) (string):find(substring) Lua: returns nil string indexOfSubCollection: substring startingAt: startpos ifAbsent: aBlock string findString: substring startingAt: startpos: Smalltalk (Squeak, Pharo) evaluate aBlock which is a block closure (or any object understanding value) returns 0
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 ...
Naively computing the hash value for the substring s[i+1..i+m] requires O(m) time because each character is examined. Since the hash computation is done on each loop, the algorithm with a naive hash computation requires O(mn) time, the same complexity as a straightforward string matching algorithm. For speed, the hash must be computed in ...