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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.
The C preprocessor (used for C and with slight differences in C++; see below) replaces all occurrences of the nine trigraph sequences in this table by their single-character equivalents before any other processing (until C23 [5]). [6] [7]
C[c] is a table that, for each character c in the alphabet, contains the number of occurrences of lexically smaller characters in the text. The function Occ(c, k) is the number of occurrences of character c in the prefix L[1..k]. Ferragina and Manzini showed [1] that it is possible to compute Occ(c, k) in constant time.
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 ...
In order to find the number of occurrences of a given string (length ) in a text (length ), [3] We use binary search against the suffix array of T {\displaystyle T} to find the starting and end position of all occurrences of P {\displaystyle P} .
The Boyer–Moore algorithm searches for occurrences of P in T by performing explicit character comparisons at different alignments. Instead of a brute-force search of all alignments (of which there are n − m + 1 {\displaystyle n-m+1} ), Boyer–Moore uses information gained by preprocessing P to skip as many alignments as possible.
We assume all the substrings have a fixed length m. A naïve way to search for k patterns is to repeat a single-pattern search taking O(n+m) time, totaling in O((n+m)k) time. In contrast, the above algorithm can find all k patterns in O(n+km) expected time, assuming that a hash table check works in O(1) expected time.
Although the substring occurrences always overlap, it is impossible to obtain a longer common substring by "uniting" them. The strings "ABABC", "BABCA" and "ABCBA" have only one longest common substring, viz. "ABC" of length 3.