Seems your visited array is visible to all recursion branch?

hi @K01egA , g . f is not equivalent to f . g
if f: A->B and g: B->A
then

f . g = id_B
g . f = id_A

which imply that for all element b in B, we have id_B(b) = b
and for all element a in A ,we have id_A(a) = a

I was confuesed for a long time with (Integral i) => i -> String,feel free to comment this line, or you will use fromIntegral apply to n

Do not mix up isInfixOf and isSubsequenceOf :)

I used to take a subset of target string and count the number of subset strings which matches src，but always get timeout problem.(subset calculations are really cost)...
After coming to the toilet, the inspiration came and it was written in one breath.