Algorithm Chunky Monkey

Algorithm Chunky Monkey

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🏁 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]]

Relevant Links

• Array.push()
• Array.slice()

💬 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.

Relevant Links

• Array.push()
• For Loops

🌻 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.

Relevant Links

• Array.push()
• Array.slice()
• For Loops

🚨 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.

Relevant Links

• Array.push()
• Array.slice()
• While Loops

🏆 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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