Oliver Eyton-Williams ee1e8abd87
feat(curriculum): restore seed + solution to Chinese (#40683)
* feat(tools): add seed/solution restore script

* chore(curriculum): remove empty sections' markers

* chore(curriculum): add seed + solution to Chinese

* chore: remove old formatter

* fix: update getChallenges

parse translated challenges separately, without reference to the source

* chore(curriculum): add dashedName to English

* chore(curriculum): add dashedName to Chinese

* refactor: remove unused challenge property 'name'

* fix: relax dashedName requirement

* fix: stray tag

Remove stray `pre` tag from challenge file.

Signed-off-by: nhcarrigan <nhcarrigan@gmail.com>

Co-authored-by: nhcarrigan <nhcarrigan@gmail.com>
2021-01-12 19:31:00 -07:00

2.9 KiB

id, title, challengeType, forumTopicId, dashedName
id title challengeType forumTopicId dashedName
594faaab4e2a8626833e9c3d Tokenize a string with escaping 5 302338 tokenize-a-string-with-escaping

--description--

Write a function or program that can split a string at each non-escaped occurrence of a separator character.

It should accept three input parameters:

  • The string
  • The separator character
  • The escape character

It should output a list of strings.

Rules for splitting:

  • The fields that were separated by the separators, become the elements of the output list.
  • Empty fields should be preserved, even at the start and end.

Rules for escaping:

  • "Escaped" means preceded by an occurrence of the escape character that is not already escaped itself.
  • When the escape character precedes a character that has no special meaning, it still counts as an escape (but does not do anything special).
  • Each occurrences of the escape character that was used to escape something, should not become part of the output.

Demonstrate that your function satisfies the following test-case:

Given the string

one^|uno||three^^^^|four^^^|^cuatro|

and using | as a separator and ^ as escape character, your function should output the following array:

  ['one|uno', '', 'three^^', 'four^|cuatro', '']

--hints--

tokenize should be a function.

assert(typeof tokenize === 'function');

tokenize should return an array.

assert(typeof tokenize('a', 'b', 'c') === 'object');

tokenize('one^|uno||three^^^^|four^^^|^cuatro|', '|', '^') should return ['one|uno', '', 'three^^', 'four^|cuatro', '']

assert.deepEqual(tokenize(testStr1, '|', '^'), res1);

tokenize('a@&bcd&ef&&@@hi', '&', '@') should return ['a&bcd', 'ef', '', '@hi']

assert.deepEqual(tokenize(testStr2, '&', '@'), res2);

--seed--

--after-user-code--

const testStr1 = 'one^|uno||three^^^^|four^^^|^cuatro|';
const res1 = ['one|uno', '', 'three^^', 'four^|cuatro', ''];

// TODO add more tests
const testStr2 = 'a@&bcd&ef&&@@hi';
const res2 = ['a&bcd', 'ef', '', '@hi'];

--seed-contents--

function tokenize(str, sep, esc) {
  return true;
}

--solutions--

// tokenize :: String -> Character -> Character -> [String]
function tokenize(str, charDelim, charEsc) {
  const dctParse = str.split('')
    .reduce((a, x) => {
      const blnEsc = a.esc;
      const blnBreak = !blnEsc && x === charDelim;
      const blnEscChar = !blnEsc && x === charEsc;

      return {
        esc: blnEscChar,
        token: blnBreak ? '' : (
          a.token + (blnEscChar ? '' : x)
        ),
        list: a.list.concat(blnBreak ? a.token : [])
      };
    }, {
      esc: false,
      token: '',
      list: []
    });

  return dctParse.list.concat(
    dctParse.token
  );
}