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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.
Longest path problem [3]: ND29 Maximum bipartite subgraph or (especially with weighted edges) maximum cut. [2] [3]: GT25, ND16 Maximum common subgraph isomorphism problem [3]: GT49 Maximum independent set [3]: GT20 Maximum Induced path [3]: GT23 Minimum maximal independent set a.k.a. minimum independent dominating set [4]
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.
Range minimum query reduced to the lowest common ancestor problem.. Given an array A[1 … n] of n objects taken from a totally ordered set, such as integers, the range minimum query RMQ A (l,r) =arg min A[k] (with 1 ≤ l ≤ k ≤ r ≤ n) returns the position of the minimal element in the specified sub-array A[l …
This list of mathematical series contains formulae for finite and infinite sums. It can be used in conjunction with other tools for evaluating sums. Here, is taken to have the value
[2] [3] There is an optimization version of the partition problem, which is to partition the multiset S into two subsets S 1, S 2 such that the difference between the sum of elements in S 1 and the sum of elements in S 2 is minimized. The optimization version is NP-hard, but can be solved efficiently in practice. [4]
In this decision problem, the input is a graph G and a number k; the desired output is yes if G contains a path of k or more edges, and no otherwise. [1] If the longest path problem could be solved in polynomial time, it could be used to solve this decision problem, by finding a longest path and then comparing its length to the number k ...
The maximum sum is 1, attained by giving one agent the item with value 1 and the other agent nothing. But the max-min allocation gives each agent value at least e, so the sum must be at most 3e. Therefore the POF is 1/(3e), which is unbounded. Alice has two items with values 1 and e, for some small e>0. George has two items with value e. The ...