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The task was taken from leetcode

Every email consists of a local name and a domain name, separated by the @ sign.

For example, in alice@leetcode.com, alice is the local name, and leetcode.com is the domain name.

Besides lowercase letters, these emails may contain '.'s or '+'s.

If you add periods ('.') between some characters in the local name part of an email address, mail sent there will be forwarded to the same address without dots in the local name. For example, "alice.z@leetcode.com" and "alicez@leetcode.com" forward to the same email address. (Note that this rule does not apply for domain names.)

If you add a plus ('+') in the local name, everything after the first plus sign will be ignored. This allows certain emails to be filtered, for example m.y+name@email.com will be forwarded to my@email.com. (Again, this rule does not apply for domain names.)

It is possible to use both of these rules at the same time.

Given a list of emails, we send one email to each address in the list. How many different addresses actually receive mails?

Example 1:

Input: ["test.email+alex@leetcode.com","test.e.mail+bob.cathy@leetcode.com","testemail+david@lee.tcode.com"]

Output: 2

Explanation: "testemail@leetcode.com" and "testemail@lee.tcode.com" actually receive mails

Note:

1 <= emails[i].length <= 100

1 <= emails.length <= 100

Each emails[i] contains exactly one '@' character.

All local and domain names are non-empty.

Local names do not start with a '+' character.

My declarative solution

/**
 * @param {string[]} emails
 * @return {number}
 */
var numUniqueEmails = emails => {
    return emails.reduce((validMails, mail) => {
      const names = mail.split('@');
      let [local, domain] = names;
      const iPlus = [...local].findIndex(x => x === '+');
      if (iPlus !== -1) { local = local.substr(0, iPlus); }
      const key = local.split('.').join('') + '@' + domain;
      if (!validMails.has(key)) { validMails.add(key); } 
      return validMails
    }, new Set).size;
};

My Imperative solution

/**
 * @param {string[]} emails
 * @return {number}
 */
var numUniqueEmails2 = emails => {
  const validMails = new Set();
  for (const mail of emails) {
    let [local, domain] = mail.split('@');
    const iPlus = [...local].findIndex(x => x === '+');
    if (iPlus !== -1) { local = local.substr(0, iPlus); }
    const key = [...local].filter(x => x !== '.').join() + '@' + domain;
    if (!validMails.has(key)) { validMails.add(key); }
  }
  return validMails.size;
};

My solution with regex

/**
 * @param {string[]} emails
 * @return {number}
 */
var numUniqueEmails3 = emails => {
  const validMails = new Set();
  for (const mail of emails) {
    let [local, domain] = mail.split('@');
    local = local.replace(/\+(.*)$/, '')
      .replace(/\./g, '');
    console.log(local);
    const key = `${local}@${domain}`;
    if (!validMails.has(key)) { validMails.add(key); }
  }
  return validMails.size;
};

Addendum

in numUniqueEmails ran the code with this snippet:

  const names = mail.split('@');
  let [local, domain] = names;

and got 90th percentile. Running it again with this snippet:

  let [local, domain] = mail.split('@');

gives me 25th percentile. I'd assume it should be the opposite. Does anyone know why this is the case and also how come the difference is so big?

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  • \$\begingroup\$ "Does anyone know why this is the case and also how come the difference is so big?" its not the code that is at fault. Leetcode's distributed processing service is to blame. The very same code submitted at different times of the day and week can return from above 90% to below 50%. It is more a metric of their service load and process spawning latency (Also likely that processes are run on variety of CPU's and clock speeds) \$\endgroup\$ – Blindman67 May 17 at 14:44
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  • I don't know what you mean by declarative in this case. All these functions are pure (except the console.log), but the implementations of the functions are all imperative (which is fine of course)
  • When adding to a set, you don't need to check if the key exists first. Just add it.
  • Splitting the string into an array of chars isn't necessary. You can use string utils instead, for example indexOf('+'). You can also do a split('+')[0] to avoid having the iPlus variable.
  • If you swap the reduce for a map, and wrap that in a Set, you don't need to add to set explicitly. It's probably a bit slower though.
  • You could add some more newlines to the code to group related concepts, and make it more readable

I came up with the solution below:

const getUniqueEmailKey = email => {
    const [local, domain] = email.split('@')
    return local
        .split('+')[0] // Take everything before +
        .split('.').join('') // Remove dots
        + '@' + domain
}

const numUniqueEmails = emails => new Set(emails.map(getUniqueEmailKey)).size

About leetcode: Try running the code a few more times. The time you get is a bit random...

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  • \$\begingroup\$ You could replace the first split with a string.slice(), from the beginning of the string to the first instance of + (you can use string.indexOf() to find it). You can replace the second split with a global regex replace, replacing all . with blank strings. This way, you don't create a lot of intermediate arrays. local.slice(0, local.indexOf('+')).replace(/\./g, '') \$\endgroup\$ – Joseph May 16 at 20:21
  • 1
    \$\begingroup\$ I don't think slice will work as expected if there are no + signs though. The regex is definitely better, should have thought of that \$\endgroup\$ – Magnus Jeffs Tovslid May 16 at 20:31

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