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Whenever the sum of the current element in the first array and the current element in the second array is more than T, the algorithm moves to the next element in the first array. If it is less than T, the algorithm moves to the next element in the second array. If two elements that sum to T are found, it stops. (The sub-problem for two elements ...
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.
Then, one can traverse the two arrays in opposite directions and enumerate all allocations in the Pareto frontier. The run-time is O ( n ⋅ c ) {\displaystyle O(n\cdot c)} . Nicosia, Pacifici and Pferschy study the price of fairness , that is, the ratio between the maximum sum of utilities, and the maximum sum of utilities in a fair solution:
A Fenwick tree or binary indexed tree (BIT) is a data structure that stores an array of values and can efficiently compute prefix sums of the values and update the values. It also supports an efficient rank-search operation for finding the longest prefix whose sum is no more than a specified value.
In the subset sum problem, the goal is to find a subset of S whose sum is a certain target number T given as input (the partition problem is the special case in which T is half the sum of S). In multiway number partitioning , there is an integer parameter k , and the goal is to decide whether S can be partitioned into k subsets of equal sum ...
The k-way merge problem consists of merging k sorted arrays to produce a single sorted array with the same elements. Denote by n the total number of elements. n is equal to the size of the output array and the sum of the sizes of the k input arrays. For simplicity, we assume that none of the input arrays is empty.
If an array index is needed to keep track of the algorithm's position in the input sequence, it doesn't change the overall constant space bound. The algorithm's bit complexity (the space it would need, for instance, on a Turing machine ) is higher, the sum of the binary logarithms of the input length and the size of the universe from which the ...
That is, it recomputes the same path costs over and over. However, we can compute it much faster in a bottom-up fashion if we store path costs in a two-dimensional array q[i, j] rather than using a function. This avoids recomputation; all the values needed for array q[i, j] are computed ahead of time only once.