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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
The longest common substrings of a set of strings can be found by building a generalized suffix tree for the strings, and then finding the deepest internal nodes which have leaf nodes from all the strings in the subtree below it. The figure on the right is the suffix tree for the strings "ABAB", "BABA" and "ABBA", padded with unique string ...
string" is a substring of "substring" In formal language theory and computer science, a substring is a contiguous sequence of characters within a string. [citation needed] For instance, "the best of" is a substring of "It was the best of times". In contrast, "Itwastimes" is a subsequence of "It was the best of times", but not a substring.
After computing E(i, j) for all i and j, we can easily find a solution to the original problem: it is the substring for which E(m, j) is minimal (m being the length of the pattern P.) Computing E(m, j) is very similar to computing the edit distance between two strings.
Comparison of two revisions of an example file, based on their longest common subsequence (black) A longest common subsequence (LCS) is the longest subsequence common to all sequences in a set of sequences (often just two sequences).
The loop at the center of the function only works for palindromes where the length is an odd number. The function works for even-length palindromes by modifying the input string. The character '|' is inserted between every character in the inputs string, and at both ends. So the input "book" becomes "|b|o|o|k|".
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 ...
The second case reduces to the first by splitting the string at the split point to create two new leaf nodes, then creating a new node that is the parent of the two component strings. For example, to split the 22-character rope pictured in Figure 2.3 into two equal component ropes of length 11, query the 12th character to locate the node K at ...