Search results
Results From The WOW.Com Content Network
rfind(string,substring) returns integer Description Returns the position of the start of the last occurrence of substring in string. If the substring is not found most of these routines return an invalid index value – -1 where indexes are 0-based, 0 where they are 1-based – or some value to be interpreted as Boolean FALSE. Related instr
If the tree is traversed from the bottom up with a bit vector telling which strings are seen below each node, the k-common substring problem can be solved in () time. If the suffix tree is prepared for constant time lowest common ancestor retrieval, it can be solved in () time. [2]
In computer science, the longest repeated substring problem is the problem of finding the longest substring of a string that occurs at least twice. This problem can be solved in linear time and space Θ ( n ) {\displaystyle \Theta (n)} by building a suffix tree for the string (with a special end-of-string symbol like '$' appended), and finding ...
T[y 2] is a substring of T with the minimal edit distance to the pattern P. Computing the E(x, y) array takes O(mn) time with the dynamic programming algorithm, while the backwards-working phase takes O(n + m) time. Another recent idea is the similarity join.
A basic example of string searching is when the pattern and the searched text are arrays of elements of an alphabet Σ. Σ may be a human language alphabet, for example, the letters A through Z and other applications may use a binary alphabet (Σ = {0,1}) or a DNA alphabet (Σ = {A,C,G,T}) in bioinformatics.
For example, the Levenshtein distance of all possible suffixes might be stored in an array , where [] [] is the distance between the last characters of string s and the last characters of string t. The table is easy to construct one row at a time starting with row 0.
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:
A naive string matching algorithm compares the given pattern against all positions in the given text. Each comparison takes time proportional to the length of the pattern, and the number of positions is proportional to the length of the text. Therefore, the worst-case time for such a method is proportional to the product of the two lengths.