Shortest Unique Substring Query Revisited

Abstract

Abstract. We revisit the problem of finding shortest unique substring (SUS) proposed recently by [6]. We propose an optimal O(n) time and space algorithm that can find an SUS for every location of a string of size n. Our algorithm significantly improves the O(n 2 ) time complexity needed by [6]. We also support finding all the SUSes covering every location, whereas the solution in [6] can find only one SUS for every location. Further, our solution is simpler and easier to implement and can also be more space efficient in practice, since we only use the inverse suffix array and longest common prefix array of the string, while the algorithm in [6] uses the suffix tree of the string and other auxiliary data structures. Our theoretical results are validated by an empirical study that shows our algorithm is much faster and more space-saving than the one in [6].

On stabbing queries for generalized longest repeat

“…In addition to the related work discussed in Section I, there were recently a sequence of work on finding shortest unique substrings (SUS) [15], [16], [17], [18], [1], of which Hu et al [1] studied the generalized version of SUS finding: Given a string position interval [x..y], 1 ≤ x ≤ y ≤ n, find SUS y x , the shortest unique substring that covers the string position interval [x..y], or the fact that such SUS y x does not exist. To the best of our knowledge, no efficient reduction from LR finding to SUS finding is known as of now.…”

“…Given the lcp and rank arrays of the string S, we can compute its useful LLRs in O(n) time and space. Proof: By Lemma 2, we know if LLR i−1 exists, the right boundary of LLR i is on or after the right boundary of LLR i−1 , for any i ≥ 2, so we can construct the array of useful LLRs in one pass as follows: we calculate each LLR i using Lemma 1, , 8), (7,13), (10,14), (11,17)}, where each useful LLR is a (start, end) tuple, representing the start and ending position of the LLR. By viewing the start and end positions as the x and y coordinates, all the useful LLRs of the example string can be visualized as the dark dots in the figure. (B) Queries for LR 12 11 , LR 14 11 , LR 12 6 and LR 5 5 are visualized by the red, blue, green, and black polylines, numbered A -D , respectively.…”

“…Search A is for LR 12 11 . That is to find all heaviest dots in S 11,12 , which include dot (7,13) and dot (11,17). Suppose the 2d DMQ launched by search A returns dot (7,13), which has a weight of 7 and is one choice for LR 12 11 .…”

Shortest Unique Queries on Strings

“…It thus remains to give a structure for finding Candidate 4. [3,6], [6,7], [7,10] We want to store I in a data structure to answer such queries efficiently.…”

“…In their initial study [9], Pei et al showed how to construct in O(n 2 ) time an index of O(n) size that answers a query in O(1) time. Soon after that, Ileri et al [6] and Tsuruta et al [10] independently improved the construction time to O(n). It is worth mentioning that O(n) size is considered optimal in the sense that D itself requires Ω(n) words to store when the alphabet is large.…”