使用正则表达式查找字符串中的可变字母组

Question

I'm doing a basic algorithm scripting challenge on FCC and I want to return true if a string in a first element of an array contains all of the letters of the string in the second element of the array, or false otherwise.

I've written some code for this. However I can't seem to pass one test:

mutation(["hello", "Hello"])

I've tried removing the global flag and have tried using constructor notation and literal notation based on recommendations from the FCC community, but to no avail.

This is the code:

function mutation(arr) {
  let patt = new RegExp("[arr.1]", "i");
  return patt.test(arr[0]);

}

mutation(["hello", "Hello"])

The function is supposed to return true instead it returns false . What is wrong with my code?

Answer 1

new RegExp("[arr.1]", "i") uses [arr.1] (literally) as the regular expression, which is why it doesn't work.

I wouldn't use a regular expression for this, it's simpler to do it directly. For instance:

function mutation(arr) {
  const lc = arr[0].toLowerCase();
  return [...arr[1].toLowerCase()].every(ch => lc.includes(ch));
}

...or to make searching the string not be a linear search:

function mutation(arr) {
  const chars = new Set([...arr[0].toLowerCase()]);
  return [...arr[1].toLowerCase()].every(ch => chars.has(ch));
}

...but if you have to use regex, you could generate an array with all the permutations of arr[1] ( ["Hello", "oHell", "loHel", ...] ) (perhaps using the answers here ) then create your regular expression using beginning/ending anchors ( ^ , $ ) and alternations ( | ) of all the permutations:

let permutations = /*...use the linked answers to create permutations...*/;
let patt = new RegExp("^(?:" + permutations.join("|") + ")$", "i");

If arr[1] may contain characters that are special in regular expressions, you'll need to escape them (perhaps using one of the answers here ):

let patt = new RegExp("^(?:" + permutations.map(escapeRegExp).join("|") + ")$", "i");

(Again: There is no built-in escapeRegExp , see the answers to the question linked above for how to create one.)