ee1e8abd87
* 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>
1.7 KiB
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]]);