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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.
Let A be the sum of the negative values and B the sum of the positive values; the number of different possible sums is at most B-A, so the total runtime is in (()). For example, if all input values are positive and bounded by some constant C , then B is at most N C , so the time required is O ( N 2 C ) {\displaystyle O(N^{2}C)} .
For separate items: the price-of-fairness of max-min fairness is unbounded. For example, suppose Alice has two items with values 1 and e, for some small e>0. George has two items with value e. The capacity is 1. The maximum sum is 1 - when Alice gets the item with value 1 and George gets nothing. But the max-min allocation gives both agents ...
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.
In number theory and computer science, the partition problem, or number partitioning, [1] is the task of deciding whether a given multiset S of positive integers can be partitioned into two subsets S 1 and S 2 such that the sum of the numbers in S 1 equals the sum of the numbers in S 2.
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 value of %phone-book{'John Doe'} is '555-1212'. The list of keys and values can be extracted using the built-in functions keys and values, respectively. So, for example, to print all the keys of a hash:
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 ...