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92 lines
1.6 KiB
Markdown
92 lines
1.6 KiB
Markdown
---
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id: 56533eb9ac21ba0edf2244e1
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title: Nesting For Loops
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challengeType: 1
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forumTopicId: 18248
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dashedName: nesting-for-loops
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---
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# --description--
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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:
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```js
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const arr = [
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[1, 2], [3, 4], [5, 6]
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];
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for (let i = 0; i < arr.length; i++) {
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for (let j = 0; j < arr[i].length; j++) {
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console.log(arr[i][j]);
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}
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}
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```
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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.
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# --instructions--
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Modify function `multiplyAll` so that it returns the product of all the numbers in the sub-arrays of `arr`.
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# --hints--
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`multiplyAll([[1], [2], [3]])` should return `6`
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```js
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assert(multiplyAll([[1], [2], [3]]) === 6);
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```
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`multiplyAll([[1, 2], [3, 4], [5, 6, 7]])` should return `5040`
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```js
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assert(
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multiplyAll([
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[1, 2],
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[3, 4],
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[5, 6, 7]
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]) === 5040
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);
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```
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`multiplyAll([[5, 1], [0.2, 4, 0.5], [3, 9]])` should return `54`
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```js
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assert(
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multiplyAll([
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[5, 1],
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[0.2, 4, 0.5],
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[3, 9]
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]) === 54
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);
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```
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# --seed--
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## --seed-contents--
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```js
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function multiplyAll(arr) {
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let product = 1;
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// Only change code below this line
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// Only change code above this line
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return product;
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}
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multiplyAll([[1, 2], [3, 4], [5, 6, 7]]);
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```
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# --solutions--
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```js
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function multiplyAll(arr) {
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let product = 1;
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for (let i = 0; i < arr.length; i++) {
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for (let j = 0; j < arr[i].length; j++) {
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product *= arr[i][j];
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}
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}
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return product;
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}
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```
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