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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
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 substring index is a data structure which gives substring search in a text or text collection in sublinear time. Once constructed from a document or set of documents, a substring index can be used to locate all occurrences of a pattern in time linear or near-linear in the pattern size, with no dependence or only logarithmic dependence on the document size.
In the array containing the E(x, y) values, we then choose the minimal value in the last row, let it be E(x 2, y 2), and follow the path of computation backwards, back to the row number 0. If the field we arrived at was E(0, y 1), then T[y 1 + 1] ... T[y 2] is a substring of T with the minimal edit distance to the pattern P.
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 informal terms, this algorithm considers every possible substring of the input string and sets [,,] to be true if the substring of length starting from can be generated from the nonterminal . Once it has considered substrings of length 1, it goes on to substrings of length 2, and so on.
Then if P is shifted to k 2 such that its left end is between c and k 1, in the next comparison phase a prefix of P must match the substring T[(k 2 - n)..k 1]. Thus if the comparisons get down to position k 1 of T, an occurrence of P can be recorded without explicitly comparing past k 1. In addition to increasing the efficiency of Boyer–Moore ...
The picture shows two strings where the problem has multiple solutions. 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.