# Algorithm Chunky Monkey  Remember to use `Read-Search-Ask` if you get stuck. Try to pair program and write your own code ### Problem Explanation:

Our goal for this Algorithm is to split `arr` (first argument) into smaller chunks of arrays with the length provided by `size` (second argument). There are 4 green checks (objectives) our code needs to pass in order to complete this Algorithm:

1. `(['a', 'b', 'c', 'd'], 2)` is expected to be `[['a', 'b'], ['c', 'd']]`
2. `([0, 1, 2, 3, 4, 5], 3)` is expected to be `[[0, 1, 2], [3, 4, 5]]`
3. `([0, 1, 2, 3, 4, 5], 2)` is expected to be `[[0, 1], [2, 3], [4, 5]]`
4. `([0, 1, 2, 3, 4, 5], 4)` is expected to be `[[0, 1, 2, 3], [4, 5]]`

## Hint: 1

The links above suggest to use `Array.push()`, so let’s start by first creating a new array to store the smaller arrays we will soon have like this:

``````var newArray = [];
``````

try to solve the problem now

## Hint: 2

Next we’ll need a `for loop` to loop through `arr`.

try to solve the problem now

## Hint: 3

Finally, we need a method to do the actual splitting and we can use `Array.slice()` to do that. The key to this Algorithm is understanding how a `for loop`, `size`, `Array.slice()` and `Array.push()` all work together.

try to solve the problem now

## Spoiler Alert! Solution ahead!

## Basic Code Solution:

``````function chunkArrayInGroups(arr, size) {

var temp = [];
var result = [];

for (var a = 0; a < arr.length; a++) {
if (a % size !== size - 1)
temp.push(arr[a]);
else {
temp.push(arr[a]);
result.push(temp);
temp = [];
}
}

if (temp.length !== 0)
result.push(temp);
return result;
}
`````` Run Code

### Code Explanation:

• Firstly, we create two empty arrays called `temp` and `result`, which we will eventually return.
• Our for loop loops until `a` is equal to or more than the length of the array in our test.
• Inside our loop, we push to `temp` using `temp.push(arr[a]);` if the remainder of `a / size` is not equal to `size - 1`.
• Otherwise, we push to `temp`, push `temp` to the `result` variable and reset `temp` to an empty array.
• Next, if `temp` isn’t an empty array, we push it to `result`.
• Finally, we return the value of `result`.

## Intermediate Code Solution:

``````function chunkArrayInGroups(arr, size) {
// Break it up
// It's already broken :(
arr = arr.slice();
var arr2 = [];
for(var i = 0, len = arr.length; i < len; i+=size) {
arr2.push(arr.slice(0, size));
arr = arr.slice(size);
}
return arr2;
}
`````` Run Code

### Code Explanation:

• Firstly, we slice `arr` using `arr.slice()` and create an empty array called `arr2`.
• Our for loop loops until `i` is equal to or more than the length of the array in our test. We also add `size` onto `i` each time we loop.
• Inside our loop, we push to `arr2` using `arr.slice(0, size)`.
• After pushing to `arr2`, we let `arr` equal to `arr.slice(size)`.
• Finally, we return the value of `arr2`.

## Advanced Code Solution:

``````function chunkArrayInGroups(arr, size) {
// Break it up.
var newArr = [];
var i = 0;

while (i < arr.length) {
newArr.push(arr.slice(i, i+size));
i += size;
}
return newArr;
}
chunkArrayInGroups(["a", "b", "c", "d"], 2);
`````` Run Code

### Code Explanation:

• Firstly, we create two variables. `newArr` is an empty array which we will push to. We also have the `i` variable set to zero, for use in our while loop.
• Our while loop loops until `i` is equal to or more than the length of the array in our test.
• Inside our loop, we push to the `newArr` array using `arr.slice(i, i+size)`. For the first time it loops, it will look something like:
``````newArr.push(arr.slice(1, 1+2))
``````
• After we push to `newArr`, we add the variable of `size` onto `i`.
• Finally, we return the value of `newArr`.

### Credits:

If you found this page useful, you can give thanks by copying and pasting this on the main chat:

`Thanks @kirah1314 @Rafase282 @jsommamtek @oshliaer for your help with Algorithm: Chunky Monkey`

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