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

1.7 KiB

id, title, challengeType, videoUrl, forumTopicId, dashedName
id title challengeType videoUrl forumTopicId dashedName
56533eb9ac21ba0edf2244e1 Nesting For Loops 1 https://scrimba.com/c/cRn6GHM 18248 nesting-for-loops

--description--

If you have a multi-dimensional array, you can use the same logic as the prior waypoint to loop through both the array and any sub-arrays. Here is an example:

var arr = [
  [1,2], [3,4], [5,6]
];
for (var i=0; i < arr.length; i++) {
  for (var j=0; j < arr[i].length; j++) {
    console.log(arr[i][j]);
  }
}

This outputs each sub-element in arr one at a time. Note that for the inner loop, we are checking the .length of arr[i], since arr[i] is itself an array.

--instructions--

Modify function multiplyAll so that it returns the product of all the numbers in the sub-arrays of arr.

--hints--

multiplyAll([[1],[2],[3]]) should return 6

assert(multiplyAll([[1], [2], [3]]) === 6);

multiplyAll([[1,2],[3,4],[5,6,7]]) should return 5040

assert(
  multiplyAll([
    [1, 2],
    [3, 4],
    [5, 6, 7]
  ]) === 5040
);

multiplyAll([[5,1],[0.2, 4, 0.5],[3, 9]]) should return 54

assert(
  multiplyAll([
    [5, 1],
    [0.2, 4, 0.5],
    [3, 9]
  ]) === 54
);

--seed--

--seed-contents--

function multiplyAll(arr) {
  var product = 1;
  // Only change code below this line

  // Only change code above this line
  return product;
}

multiplyAll([[1,2],[3,4],[5,6,7]]);

--solutions--

function multiplyAll(arr) {
  var product = 1;
  for (var i = 0; i < arr.length; i++) {
    for (var j = 0; j < arr[i].length; j++) {
      product *= arr[i][j];
    }
  }
  return product;
}

multiplyAll([[1,2],[3,4],[5,6,7]]);