Determine if string has at least 2 same elements from an array

Question

I want to determine if string has at least 2 same elements from the array

const array = ["!", "?"];

const string1 = "!hello"; // should return false
const string2 = "!hello?"; // should return false
const string3 = "!hello!"; // should return true
const string4 = "hello ??"; // should return true
const string5 = "hello ?test? foo"; // should return true
const string6 = "hello ?test ?? foo"; // should return true

I'm not sure what is gonna be better: a regex or a function? Any would be fine.

I tried this:

const array = ["!", "?"];
const string = "test!";

array.every(ar => !string.includes(ar));

But it only detects if there at least 1 elements from array, not 2.

Answer 1

You can use Array#some and String#split to do it:

const check=(array,string)=>array.some(char=>(string.split(char).length-1)>=2)

const array = ["!", "?"];

console.log(check(array,"!hello"))
console.log(check(array,"!hello?"))
console.log(check(array,"!hello!"))
console.log(check(array,"hello ??"))
console.log(check(array,"hello ?test? foo"))
console.log(check(array, "hello ?test ?? foo"))

How does it work?

Let's split up (I mean to split() up)!

const check=(array,string)=>
  array.some(char=>
    (
      string.split(char)
      .length-1
    )>=2
  )

• First, use Array#some, which tests that at least one element of the array should pass (i.e. either ? or !)
• Split up the string by char, and count how many parts do we have
  • If we have n parts, it means that we have n-1 places where the char matches. (e.g. 2 | splits a string into 3 parts: a|b|c)
• Finally, test whether we have 2 or more delimiters