I got this code:

`let a = "ABCDEFGHIJ";;`

What is the best and efficient way to calculate hash values for all substrings in a string?

Some substrings of string “a”.

A -> 65

AB -> 23

BC -> 324

ABC -> 3234

…

I got this code:

`let a = "ABCDEFGHIJ";;`

What is the best and efficient way to calculate hash values for all substrings in a string?

Some substrings of string “a”.

A -> 65

AB -> 23

BC -> 324

ABC -> 3234

…

Since the hash values are shown in the examples, is the hash function already provided as well?

Let me see if I can answer this. Looking at the problem, since you need to calculate “all substrings”, you will end up with at least an O(n^2) time complexity for getting the start and end indexes of each substring (with a simple nested loop), because there will be n^2 substrings for a string of length n.

After we get the indexes, the next step is to calculate the hash function. A naive approach would be to calculate the hash for the whole substring in each iteration, which may give us an O(n) time complexity if we need to go through all the characters of the substring from start to end indexes again. This gives the overall time complexity of O(n^3), and O(n^2) space complexity to store the final result.

We may be able to further optimize this if the hash function is such that for a string “ABCD”, the value of `hash("ABCD")`

can be achieved by doing constant math calculation of `hash("ABC")`

and `hash("D")`

(i.e., the hash value of a string is calculated from the hash value of the substring up to the second last character and and the hash value of the last character, or `let hash x = f(hash(x[start:end-1]), hash(x[end]))`

for a constant function `f`

).

With this, we can calculate the hash values constantly during the nested loop, store the result in some kind of map with start and end index as key, and use that result for the next calculation, which gives us the final time complexity of O(n^2). It still has the O(n^2) space complexity for the final result and the temporary map.

The hash function may be derivable from the examples, e.g. since `A -> 65`

then it might be related to a calculation with the ASCII representation of the characters.

I need some function more efficient than Hashtbl.hash “ABCDE”;;. Some function using lxor

https://en.wikipedia.org/wiki/Rolling_hash might be relevant, though I’m not hugely familiar with it.

The string only have chars ‘A’, ‘C’, ‘T’ and’G’. Example AATG

I try this code:

`let a = 1;; let c = 2;; let g = 3;; let t = 4;; let base = 100003;; let f1((str:string)) : int = let hs = ref 0 in let lstr = String.length str in for i = 0 to lstr-1 do if String.get str i = 'A' then hs := (!hs + a) mod lstr else if String.get str i = 'C' then hs := (!hs + c) mod lstr else if String.get str i = 'G' then hs := (!hs + g) mod lstr else hs := (!hs + t) mod lstr done; !hs; ;;`

but for the strings with length high (>100000), this function f1 it’s slower than Hashtbl.hash “SameString”;;. Any suggestions?

Since Hashtbl.hash is highly optimized C, it’s almost impossible to write a more efficient hash function in OCaml without using some kind of rolling hash or @bobbypriambodo’s suggestion.

The bottleneck in your hash function is probably the `mod`

, which is a slow operation, and maybe the fact that you’re repeatedly doing `String.get str i`

for the same `i`

.

Apart from that, your hash function is very weak. All permutations of a string (e.g. ACGA, GACA, AACG, CGAA, CAAG, …) will hash to the same value.