Given a sequence of n real numbers and an integer k, 1 k 1 2 n(n − 1), the k maximum-sum segments problem is to locate the k segments whose sums are the k largest among all possible segment sums. Recently, Bengtsson and Chen gave an O(min{k + n log 2 n, n √ k})-time algorithm for this problem. Bae and Takaoka later proposed a more efficient algorithm for small k. In this paper, we propose an O(n + k log(min{n, k}))-time algorithm for the same problem, which is superior to both of them when k is o(n log n). We also give the first optimal algorithm for delivering the k maximum-sum segments in non-decreasing order if k n. Then we develop an O(n 2d−1 +k log min{n, k})-time algorithm for the d-dimensional version of the problem, where d > 1 and each dimension, without loss of generality, is of the same size n. This improves the best previously known O(n 2d−1 C)-time algorithm, also by Bengtsson and Chen, where C = min{k + n log 2 n, n √ k}. It should be pointed out that, given a two-dimensional array of size m×n, our algorithm for finding the k maximum-sum subarrays is the first one achieving cubic time provided that k is O(m 2 n/ log n).

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Given a sequence of n real numbers and an integer k, 1 k 1 2 n(n − 1), the k maximum-sum segments problem is to locate the k segments whose sums are the k largest among all possible segment sums. Recently, Bengtsson and Chen gave an O(min{k + n log 2 n, n √ k})-time algorithm for this problem. Bae and Takaoka later proposed a more efficient algorithm for small k. In this paper, we propose an O(n + k log(min{n, k}))-time algorithm for the same problem, which is superior to both of them when k is o(n log n). We also give the first optimal algorithm for delivering the k maximum-sum segments in non-decreasing order if k n. Then we develop an O(n 2d−1 +k log min{n, k})-time algorithm for the d-dimensional version of the problem, where d > 1 and each dimension, without loss of generality, is of the same size n. This improves the best previously known O(n 2d−1 C)-time algorithm, also by Bengtsson and Chen, where C = min{k + n log 2 n, n √ k}. It should be pointed out that, given a two-dimensional array of size m×n, our algorithm for finding the k maximum-sum subarrays is the first one achieving cubic time provided that k is O(m 2 n/ log n).