# ordered set inclusion 2

Slightly change the previous question, ordered set inclusion,

Now, consider $$A=\{2,3\}$$ and $$B=\{2,3,4,5\}, C=\{5,2,4,3\}$$, Now I want to make element $$\{2,3\}$$ is in $$B$$ as well as $$C$$, what can be useful command for mathematica? [Here $$\{3,2\}$$ is not in $$B$$, $$C$$]

For example, given (for example) $$A=\{ 3,4,5\}$$ and set $$B=\{9,8,3,2,4,7,5\}$$, want to make function which produce true. [Since the order $$3,4,5$$ is same with B.

ClearAll[orderedSubsetQ]
orderedSubsetQ = MemberQ[Subsets[#2, {Length @ #}], #] &;


Examples:

a = {2, 3} ; b = {2, 3, 4, 5} ; c = {5, 2, 4, 3};

orderedSubsetQ[a, b]

True

orderedSubsetQ[a, c]

True

orderedSubsetQ[{3,2}, b]

False

orderedSubsetQ[{3,2}, c]

False

orderedSubsetQ[{3, 4, 5}, {9, 8, 3, 2, 4, 7, 5}]

 True

orderedSubsetQ[{4, 3, 5}, {9, 8, 3, 2, 4, 7, 5}]

 False


You can also use

ClearAll[orderedSubsetQ2, orderedSubsetQ3, orderedSubsetQ4, orderedSubsetQ5]

orderedSubsetQ2 = MatchQ[#2, Riffle[#, ___, {1, -1, 2}]] &;

orderedSubsetQ3 = DeleteCases[#2, Except[Alternatives @@ #]] == # &;

orderedSubsetQ4 = SequenceCases[#2, Riffle[#, ___]] =!= {} &;

orderedSubsetQ5 = Positive @ SequenceCount[#2, Riffle[#, ___]] &;