The non-overlapping indexing problem is defined as follows: pre-process a given text $\mathsf{T}[1,n]$ of length n into a data structure such that whenever a pattern $P[1,m]$ comes as an input, we can efficiently report the largest set of non-overlapping occurrences of P in $\mathsf{T}$. The best-known solution is by Cohen and Porat [ISAAC 2009]. The size of their structure is $O(n)$ words and the query time is optimal $O(m+\mathsf{nocc})$, where $\mathsf{nocc}$ is the output size. Later, Ganguly et al. [CPM 2015 and Algorithmica 2020] proposed a compressed space solution. We study this problem in the cache-oblivious model and present a new data structure of size $O(n\log n)$ words. It can answer queries in optimal $O(\frac{m}{B}+{\log }_{B}n+\frac{\mathsf{nocc}}{B})$ I/O operations, where B is the block size. The space can be improved to $O(n{\log }_{M/B}n)$ in the cache-aware model, where M is the size of main memory. Additionally, we study a generalization of this problem with an additional range $[s,e]$ constraint. Here the task is to report the largest set of non-overlapping occurrences of P in $\mathsf{T}$, that are within the range $[s,e]$. We present an $O(n{\log }^{2}n)$ space data structure in the cache-aware model that can answer queries in optimal $O(\frac{m}{B}+{\log }_{B}n+\frac{{\mathsf{nocc}}_{[s,e]}}{B})$ I/O operations, where ${\mathsf{nocc}}_{[s,e]}$ is the output size.

非重叠索引问题定义如下：预处理给定文本 $\mathsf{Ť}[\mathrm{1个}，ñ]$长度为n的数据结构$P[\mathrm{1个}，米]$作为输入，我们可以有效地报告P中最大的一组非重叠出现$\mathsf{Ť}$。最著名的解决方案是Cohen和Porat [ISAAC 2009]。其结构的大小是$Ø（ñ）$ 单词和查询时间最佳 $Ø（米+\mathsf{Nocc}）$，在哪里 $\mathsf{Nocc}$是输出大小。后来，Ganguly等人。[CPM 2015和Algorithmica 2020]提出了一种压缩空间解决方案。我们在忽略缓存的模型中研究了此问题，并提出了一个新的大小数据结构$Ø（ñ\mathrm{日志}ñ）$话。它可以最佳地回答查询$Ø（\frac{米}{乙}+{\mathrm{日志}}_{乙}ñ+\frac{\mathsf{Nocc}}{乙}）$I / O操作，其中B是块大小。空间可以改善为$Ø（ñ{\mathrm{日志}}_{\mathrm{中号}/乙}ñ）$在支持缓存的模型中，其中M是主内存的大小。此外，我们还研究了此问题的一般性$[s，Ë]$约束。这里的任务是报告P in中最大的一组非重叠出现$\mathsf{Ť}$，在范围内 $[s，Ë]$。我们提出一个$Ø（ñ{\mathrm{日志}}^2ñ）$ 缓存感知模型中的空间数据结构，可以以最佳方式回答查询 $Ø（\frac{米}{乙}+{\mathrm{日志}}_{乙}ñ+\frac{{\mathsf{Nocc}}_{[s，Ë]}}{乙}）$ I / O操作 ${\mathsf{Nocc}}_{[s，Ë]}$ 是输出大小。