Abstract. The maximum subarray problem for a one-or two-dimensional array is to find the array portion that maiximizes the sum of array elements in it. The K-maximum subarray problem is to find the K subarrays with largest sums. We improve the time complexity for the one-dimensional case from O(min{K + n log 2 n, n √
The maximum subarray problem is to find the contiguous array elements having the largest possible sum. We extend this problem to find K maximum subarrays. For general K maximum subarrays where overlapping is allowed, Bengtsson and Chen presented OðminfK + n log 2 n‚ n ffiffiffi ffi K p gÞ time algorithm for one-dimensional case, which finds unsorted subarrays. Our algorithm finds K maximum subarrays in sorted order with improved complexity of O ((n + K ) log K ). For the twodimensional case, we introduce two techniques that establish O(n 3 ) and subcubic time.
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.
Abstract. We design a faster algorithm for the k-maximum sub-array problem under the conventional RAM model, based on distance matrix multiplication (DMM). Specifically we achieve O(n 3 log log n/ log n + k log n) for a general problem where overlapping is allowed for solution arrays. This complexity is sub-cubic when k = o(n 3 / log n). The best known complexities of this problem are O(n 3 + k log n), which is cubic when k = O(n 3 / log n), and O(kn 3 log log n/ log n), which is sub-cubic when k = o( log n/ log log n).
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