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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.
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]
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
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.
Each substring is terminated with special character $. The six paths from the root to the leaves (shown as boxes) correspond to the six suffixes A$, NA$, ANA$, NANA$, ANANA$ and BANANA$. The numbers in the leaves give the start position of the corresponding suffix. Suffix links, drawn dashed, are used during construction.
A simple and inefficient way to see where one string occurs inside another is to check at each index, one by one. First, we see if there is a copy of the needle starting at the first character of the haystack; if not, we look to see if there's a copy of the needle starting at the second character of the haystack, and so forth.
In computer science, the longest repeated substring problem is the problem of finding the longest substring of a string that occurs at least twice. This problem can be solved in linear time and space Θ ( n ) {\displaystyle \Theta (n)} by building a suffix tree for the string (with a special end-of-string symbol like '$' appended), and finding ...
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 ...