Message from JavaScript talks
May 2017
— That check above never runs if these don't pass:
// Deduce if the String is definitely not in the Set from basic data.
if (charArray.length !== 0
&& uniquesLength !== 0
&& listLength !== 0
&& list.hasOwnProperty(charArray[0])
&& list[charArray[0]].hasOwnProperty([charArray.length])
) {
For "Hello", it checks for this type of path in the set:
{
H: {
5: {
// Word may exist in here
}
}
}
— So as you can probably tell, construction of the set dictates where in the set we conduct the search
— Of course if you have a set with only words which start with H, and are of all the same length, it will be only as effective as a linear search on a flat array
— But for diverse datasets, this will improve search efficiency massively
— You could also target by multiple letters, instead of just the first. I'm not sure how much more effective that would be
— All these checks are ordered from "least expensive" to "most expensive" too, so if we can deduce that the string is not in the set from a less expensive method, we just saved time :D
— Put in js file, and module.exports the class
— Should follow this api:
s = new Set()
s.add(str)
s.has(str)
s.delete(str)
— Sure
— Made the test better
— I am eager to see if it is slower or faster