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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 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.
The final result is that the last cell contains all the longest subsequences common to (AGCAT) and (GAC); these are (AC), (GC), and (GA). The table also shows the longest common subsequences for every possible pair of prefixes. For example, for (AGC) and (GA), the longest common subsequence are (A) and (G).
In computer science, the longest common prefix array (LCP array) is an auxiliary data structure to the suffix array. It stores the lengths of the longest common prefixes (LCPs) between all pairs of consecutive suffixes in a sorted suffix array. For example, if A := [aab, ab, abaab, b, baab] is a suffix array, the longest common prefix between A ...
The length of the longest decreasing subsequence is equal to the length of the first column of P. Now, it is not possible to fit ( r − 1)( s − 1) + 1 entries in a square box of size ( r − 1)( s − 1), so that either the first row is of length at least r or the last row is of length at least s .
The set ret can be saved efficiently by just storing the index i, which is the last character of the longest common substring (of size z) instead of S[(i-z+1)..i]. Thus all the longest common substrings would be, for each i in ret, S[(ret[i]-z)..(ret[i])]. The following tricks can be used to reduce the memory usage of an implementation:
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