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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 problem of computing a longest common subsequence has been well studied in computer science. It can be solved in polynomial time by dynamic programming ; [ 5 ] this basic algorithm has additional speedups for small alphabets (the Method of Four Russians ), [ 6 ] for strings with few differences, [ 7 ] for strings with few matching pairs of ...
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 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 first phase of patience sort, the card game simulation, can be implemented to take O(n log n) comparisons in the worst case for an n-element input array: there will be at most n piles, and by construction, the top cards of the piles form an increasing sequence from left to right, so the desired pile can be found by binary search. [1]
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.
For r = 3 and s = 2, the formula tells us that any permutation of three numbers has an increasing subsequence of length three or a decreasing subsequence of length two. Among the six permutations of the numbers 1,2,3: 1,2,3 has an increasing subsequence consisting of all three numbers; 1,3,2 has a decreasing subsequence 3,2
A Reed–Solomon code (like any MDS code) is able to correct twice as many erasures as errors, and any combination of errors and erasures can be corrected as long as the relation 2E + S ≤ n − k is satisfied, where is the number of errors and is the number of erasures in the block.