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If the array contains all non-negative numbers, then the problem is trivial; a maximum subarray is the entire array. If the array contains all non-positive numbers, then a solution is any subarray of size 1 containing the maximal value of the array (or the empty subarray, if it is permitted).
For shared items: define a 2-dimensional array such that (,) = iff there exists a solution giving a total weight of w i to agent i. It is possible to enumerate all possible utility profiles in time O ( n ⋅ c 2 ) {\displaystyle O(n\cdot c^{2})} where n is the number of items and c is the maximum size of an item.
The problem is NP-hard even when all input integers are positive (and the target-sum T is a part of the input). This can be proved by a direct reduction from 3SAT. [2] It can also be proved by reduction from 3-dimensional matching (3DM): [3] We are given an instance of 3DM, where the vertex sets are W, X, Y.
The loop at the center of the function only works for palindromes where the length is an odd number. The function works for even-length palindromes by modifying the input string. The character '|' is inserted between every character in the inputs string, and at both ends. So the input "book" becomes "|b|o|o|k|". The even-length palindrome "oo ...
The number of perfect matchings of the complete graph K n (with n even) is given by the double factorial (n – 1)!!. [12] The crossing numbers up to K 27 are known, with K 28 requiring either 7233 or 7234 crossings. Further values are collected by the Rectilinear Crossing Number project. [13] Rectilinear Crossing numbers for K n are
The Nial example of the inner product of two arrays can be implemented using the native matrix multiplication operator. If a is a row vector of size [1 n] and b is a corresponding column vector of size [n 1]. a * b; By contrast, the entrywise product is implemented as: a .* b;
A min-max heap is a complete binary tree containing alternating min (or even) and max (or odd) levels. Even levels are for example 0, 2, 4, etc, and odd levels are respectively 1, 3, 5, etc. We assume in the next points that the root element is at the first level, i.e., 0. Example of Min-max heap. Each node in a min-max heap has a data member ...
When the length decreases, the sequences must have had a common element. Several paths are possible when two arrows are shown in a cell. Below is the table for such an analysis, with numbers colored in cells where the length is about to decrease. The bold numbers trace out the sequence, (GA). [6]