将两个字符串与动态空白区域进行比较的最快方法？

Question

I have two strings bigstring and smallstring , and each string is a paragraph of words. However in between each word is a bunch of whitespace ( \\s in regex) characters of random length.

So for example bigstring could be like hello world . And this goes for smallstring too.

What I want to be able to do is, check if smallstring is a substring of bigstring (word for word) where the \\s+ part of it is considered the same, and case insensitively . So for example if

bigstring = "hello \\t\\r\\n world \\n foobar"

smallstring = "HELLO \\t world"

then smallstring is a substring of bigstring .

bigstring = "hello \\t\\r\\n world \\n foobar"

smallstring = "HEL"

This is not a substring (word for word), because there is no word called hel in bigstring .

bigstring = "the \\t\\r\\n nest"

smallstring = "then \\n est"

This is also not a substring (word for word).

One method is to tokenize both strings into arrays, so break up the stuff between \\s+ into tokens, and the \\s+ is the delimiters. Then literally check if one array is contained in the other array in order and consecutively with case insensitively.

However in this case, I need speed to be the priority, as it should be the fastest way.

Does anyone know a way to check this?

I was perhaps thinking of a way to check these strings as you loop through both, character by character, but not sure how to do that?

Thanks

Answer 1

I am not sure where this ranks on speed, but does this achieve your goal (now edited for edge case of 'impl' vs. 'mpl', by adding leading space)

var isSubstring = function(bigstring, smallstring) {
  bigstring = " " + bigstring.replace(/\s+/g, " ").toLowerCase() + " "
  smallstring = " " + smallstring.replace(/\s+/g, " ").toLowerCase() + " "
  return(bigstring.indexOf(smallstring) >= 0)
}

Adding a trailing (and, now, leading) space covers the case where smallstring is a single word fragment ('hel' vs. 'hello' and 'impl' vs. 'mpl' in your example above and in comments below)

Use cases:

bigstring = "hello   \t\r\n  world \n foobar"
smallstring = "HELLO \t world"
console.log(isSubstring(bigstring, smallstring))
//evaluates to true

bigstring = "hello   \t\r\n  world \n foobar"
smallstring = "HEL"
console.log(isSubstring(bigstring, smallstring))
// evaluates to false

bigstring = "impl"
smallstring = "mpl"
console.log(isSubstring(bigstring, smallstring))
// evaluates to false

Answer 2

RegExp is definitely not the fastest, but you can search the big string with a RegExp generated from the small string:

 bigstring = "hello \\t\\r\\n world \\n foobar" smallstring = "HELLO \\t world" r = new RegExp( '\\\\b' + smallstring.replace(/\\s+/g, '\\\\s+') + '\\\\b', 'i' ) console.log( r.test(bigstring), r ) // true /\\bHELLO\\s+world\\b/i 

A faster case-insensitive string search would most likely use charCodeAt and/or some kind of a word/token lookup structure, as for example https://github.com/bvaughn/js-search seems to use.

Answer 3

Let F(a) will return unified version of string a . By unified I mean that all consecutive space characters will be replaced by a single space and all letters will be moved to lower case. This function can be calculated in linear time - O(|a|) .

In this case you need to check if F(smallstring) is substring of F(bigstring) . To handle this quickly you can use some standard algo like KMP .