Algorithms for K-Disjoint Maximum Subarrays

Abstract

Abstract. The maximum subarray problem is to find the array portion that maximizes the sum of array elements in it. For K disjoint maximum subarrays, Ruzzo and Tompa gave an O(n) time solution for one-dimension. This solution is, however, difficult to extend to twodimensions. While a trivial solution of O(Kn 3 ) time is easily obtainable for two-dimensions, little study has been undertaken to better this. We first propose an O(n + K log K) time solution for one-dimension. This is equivalent to Ruzzo and Tompa's when order is considered. Based on this, we achieve O(n 3 +Kn 2 log n) time for two-dimensions. This is cubic time when K ≤ n/ log n.

EFFECT OF CORNER INFORMATION IN SIMULTANEOUS PLACEMENT OF k RECTANGLES AND TABLEAUX

“…Bae and Takaoka [4] also studied k-disjoint maximum subarrays problem in a two-dimensional array. The problem is defined in a greedy fashion such that we find the maximum weight rectangle, then find the maximum weight rectangle in the remaining part, and so on.…”

A Sub-cubic Time Algorithm for the k-Maximum Subarray Problem

“…Once the maximum subarray is found, we need to exclude the occupied portion from further considerations. This was done by "hole creation" in [3], achieving a cubic time for k = O(n/ log n). A "hole" causes many tournaments to be updated to offer the best subarrays to be chosen.…”

“…Note that we do not actually sort the set. The leftmost element of c ij , that is, the minimum, participates in the tournament for "min" in (3). If c ij = (x 1 , x 2 , ..., x l ) and x 1 is chosen as the winner, c ij is changed to (x 2 , ..., x l , ∞), etc.…”

“…The best known results for the disjoint case are the straightforward O(kM (n)), which is sub-cubic for small k such as k = o( log n/ log log n), where M (n) is the time for DMM, and O(n 3 + kn 2 log n) by Bae and Takaoka [3] for larger k. The problem here is to find the maximum, the second maximum, etc. from the remaining portion.…”