Counting distinct (non-)crossing substrings

TL;DR

This paper introduces efficient algorithms to count distinct crossing and non-crossing substrings in a string for all positions, extending previous solutions from constant alphabets to general and linearly sortable alphabets.

Contribution

It presents new algorithms that compute the counts in linear total time for all positions in the string, improving over previous methods limited to constant alphabets.

Findings

Algorithms run in O(n) total time for all positions.

Extends previous solutions to general ordered alphabets.

Provides efficient counting of crossing and non-crossing substrings.

Abstract

Let $w$ be a string of length $n$. The problem of counting factors crossing a position - Problem 64 from the textbook ``125 Problems in Text Algorithms'' [Crochemore, Leqroc, and Rytter, 2021], asks to count the number $\mathcal{C}(w,k)$ (resp. $\mathcal{N}(w,k)$) of distinct substrings in $w$ that have occurrences containing (resp. not containing) a position $k$ in $w$. The solutions provided in their textbook compute $\mathcal{C}(w,k)$ and $\mathcal{N}(w,k)$ in $O(n)$ time for a single position $k$ in $w$, and thus a direct application would require $O(n^2)$ time for all positions $k = 1, \ldots, n$ in $w$. Their solution is designed for constant-size alphabets. In this paper, we present new algorithms which compute $\mathcal{C}(w,k)$ in $O(n)$ total time for general ordered alphabets, and $\mathcal{N}(w,k)$ in $O(n)$ total time for linearly sortable alphabets, for all positions $k = 1, \ldots, n$ in $w$.