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Your problem appears to be: Given strings $S,T$, determines whether $S$ appears as a subsequence of $T$. There are several ways you could approach this. One way is to use dynamic programming and …

Let $S,T$ be two sets of size $n$. Suppose we hash each to a $m$-bit Bloom filter, using $k$ hash functions; let $x_S$ be the $m$-bit vector corresponding to $S$, and $x_T$ the $m$-bit vector corresp …

One approach is to build a Levenshtein automaton for the fixed string (see, e.g., here). Given a string $x$ and a distance $D$, you can build a DFA that recognizes all strings that are at distance $\ …

There are many possible distance metrics, and without any criteria, we have no basis to choose. Here are two plausible ones. Let $L_1,L_2$ be the languages of the two DFAs. Let $L$ be the symmetric …

This is a straightforward consequence of the fact that the edit script only contains insert and delete operations. It follows from a little bit of basic arithmetic -- it's nothing especially deep. C …

There are two considerations: running time, and correctness. Running time: When $k < (n-1)/\lg(4n)$, heuristically I expect the running time of your algorithm to be fine and I'd guess you won't experi …

Yes. This can be done, using Levenshtein automata. Let $S_k = \{y \in \{0,1\}^* : d(x,y) \le k)\}$. Then the set $S_k$ is regular, and one can construct a finite-state automaton for it, called a Le …

There's a "trick" you can use that might potentially speed up your algorithm a little: shingling. No guarantees that it'll necessarily help in your particular case, though. Lemma. If the edit distan …

If you want the minimum number of single-character changes (insert, delete, substitute) that transforms the shorter string $S$ into some length-$m$ substring of $T$, then this can be done in $O(nm)$ t …

Comparing two strings Algorithms for computing the Levenshtein edit distance between a pair of strings can be found in the Wikipedia page on the edit distance. As that page explains, the running tim …

I think you should be able to adapt the standard dynamic programming algorithm to this, with a minor modification: you simply treat deletions at the beginning and end of s2 as costing 0.