Filter array with multiple startsWith and endsWith arguments

Question

Browsed several different questions/answers surrounding this question, but most rely on includes or indexOf

Problem: How any array to filter (names in this case). I need to filter it using 2 different arrays, one of which has startsWith criteria and the other has endsWith criteria

var names = ['BOB','CATHY','JAKOB','AARON','JUSTICE','BARBARA','DANIEL','BOBBY','JUSTINE','CADEN','URI','JAYDEN','JULIE']
startPatterns = ['BO','JU']
endPatterns = ['EN','ICE']

//res = ['BOB','JUSTICE','JUSTINE','JULIE','JAYDEN','JUSTICE']

Obviously you cannot do names.filter(d => d.startsWith(startPatterns)) because startPatterns is not a string but a array. Something like this isn't working and is terrible slow too:

res=[]
names.forEach(d => {
  endPatterns.forEach(y => d.endsWith(y) ? res.push(d) : '')
  startPatterns.forEach(s => d.startsWith(s) ? res.push(d) : '')})
console.log(res)

Answer 1

You can use Array.prototype.some on the pattern arrays to achieve this:

let filtered = names.filter(name => (
  startPatterns.some(pattern => name.startsWith(pattern)) ||
  endPatterns.some(pattern => name.endsWith(pattern))
))

The logic here being "Return true if the name begins with at least one of the start patterns OR ends with at least one of the end patterns".

Answer 2

You could build a regular expression and check the string against the pattern.

var names = ['BOB','CATHY','JAKOB','AARON','JUSTICE','BARBARA','DANIEL','BOBBY','JUSTINE','CADEN','URI','JAYDEN','JULIE'],
    startPatterns = ['BO','JU'],
    endPatterns = ['EN','ICE'],
    regexp = new RegExp(`^(${startPatterns.join('|')})|(${endPatterns.join('|')})$`),
    result = names.filter(s => regexp.test(s));

console.log(result);

A non regular expression approach with an array with the methods and wanted values.

var names = ['BOB','CATHY','JAKOB','AARON','JUSTICE','BARBARA','DANIEL','BOBBY','JUSTINE','CADEN','URI','JAYDEN','JULIE'],
    patterns = [['startsWith', ['BO','JU']], ['endsWith', ['EN','ICE']]],
    result = names.filter(s => patterns.some(([k, values]) => values.some(v => s[k](v))));

console.log(result);