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The longest alternating subsequence problem has also been studied in the setting of online algorithms, in which the elements of are presented in an online fashion, and a decision maker needs to decide whether to include or exclude each element at the time it is first presented, without any knowledge of the elements that will be presented in the future, and without the possibility of recalling ...
The array L stores the length of the longest common suffix of the prefixes S[1..i] and T[1..j] which end at position i and j, respectively. The variable z is used to hold the length of the longest common substring found so far. The set ret is used to hold the set of strings which are of length z.
For LCS(R 2, C 1), A is compared with A. The two elements match, so A is appended to ε, giving (A). For LCS(R 2, C 2), A and G do not match, so the longest of LCS(R 1, C 2), which is (G), and LCS(R 2, C 1), which is (A), is used. In this case, they each contain one element, so this LCS is given two subsequences: (A) and (G).
The above algorithm has worst-case time and space complexities of O(mn) (see big O notation), where m is the number of elements in sequence A and n is the number of elements in sequence B. The Hunt–Szymanski algorithm modifies this algorithm to have a worst-case time complexity of O ( mn log m ) and space complexity of O ( mn ) , though it ...
The longest increasing subsequence problem is closely related to the longest common subsequence problem, which has a quadratic time dynamic programming solution: the longest increasing subsequence of a sequence is the longest common subsequence of and , where is the result of sorting.
For example, for the array of values [−2, 1, −3, 4, −1, 2, 1, −5, 4], the contiguous subarray with the largest sum is [4, −1, 2, 1], with sum 6. Some properties of this problem are: If the array contains all non-negative numbers, then the problem is trivial; a maximum subarray is the entire array.
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 ...
SEQ 1 = A CG G T G TCG T GCTATGCT GA T G CT G ACTTAT A T G CTA SEQ 2 = CGTTCGGCTAT C G TA C G TTCTA TT CT A T G ATT T CTA A. Another way to show this is to align the two sequences, that is, to position elements of the longest common subsequence in a same column (indicated by the vertical bar) and to introduce a special character (here, a dash ...