空间高效的STR-IC-LCS计算

Yuuki Yonemoto, Yuto Nakashima, Shunsuke Inenaga, H. Bannai
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引用次数: 1

摘要

比较两个给定字符串A和B的最基本方法之一是最长公共子序列(LCS),其中的任务是找到(长度)最长公共子序列。本文研究的STR-IC-LCS问题是Chen和Chao [J]提出的约束LCS问题之一。梳子。Optim, 2011]。如果Z是包含P作为子字符串的A和B的最长公共子序列之一,则称字符串Z是三个给定字符串A, B和P的STR-IC-LCS。我们提出了一种空间高效的STR-IC-LCS问题解决方案。我们的算法在O (n2)时间和O ((cid:96) + 1)(n−(cid:96) + 1))空间中计算STR-IC-LCS的长度,其中(cid:96)是长度为n的a和B的最长公共子序列的长度。当(cid:96) = O(1)或n - (cid:96) = O(1)时,则我们的算法仅使用线性O (n)空间。
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Space-Efficient STR-IC-LCS Computation
One of the most fundamental method for comparing two given strings A and B is the longest common subsequence (LCS), where the task is to find (the length) of the longest common subsequence. In this paper, we address the STR-IC-LCS problem which is one of the constrained LCS problems proposed by Chen and Chao [J. Comb. Optim, 2011]. A string Z is said to be an STR-IC-LCS of three given strings A , B , and P , if Z is one of the longest common subsequences of A and B that contains P as a substring. We present a space efficient solution for the STR-IC-LCS problem. Our algorithm computes the length of an STR-IC-LCS in O ( n 2 ) time and O (( (cid:96) + 1)( n − (cid:96) + 1)) space where (cid:96) is the length of a longest common subsequence of A and B of length n . When (cid:96) = O (1) or n − (cid:96) = O (1), then our algorithm uses only linear O ( n ) space.
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