* fix: restructure certifications guide articles * fix: added 3 dashes line before prob expl * fix: added 3 dashes line before hints * fix: added 3 dashes line before solutions
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title
| title |
|---|
| Chunky Monkey |
Chunky Monkey
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:
(['a', 'b', 'c', 'd'], 2)is expected to be[['a', 'b'], ['c', 'd']]([0, 1, 2, 3, 4, 5], 3)is expected to be[[0, 1, 2], [3, 4, 5]]([0, 1, 2, 3, 4, 5], 2)is expected to be[[0, 1], [2, 3], [4, 5]]([0, 1, 2, 3, 4, 5], 4)is expected to be[[0, 1, 2, 3], [4, 5]]
Relevant Links
Hints
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 = [];
Hint 2
Next we'll need a for loop to loop through arr.
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.
Solutions
Solution 1 (Click to Show/Hide)
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;
}
Code Explanation
- Firstly, we create two empty arrays called
tempandresult, which we will eventually return. - Our for loop loops until
ais equal to or more than the length of the array in our test. - Inside our loop, we push to
tempusingtemp.push(arr[a]);if the remainder ofa / sizeis not equal tosize - 1. - Otherwise, we push to
temp, pushtempto theresultvariable and resettempto an empty array. - Next, if
tempisn't an empty array, we push it toresult. - Finally, we return the value of
result.
Relevant Links
Solution 2 (Click to Show/Hide)
function chunkArrayInGroups(arr, size) {
// Break it up.
var arr2 = [];
for (var i = 0; i < arr.length; i += size) {
arr2.push(arr.slice(i, i + size));
}
return arr2;
}
Code Explanation
- First, we create an empty array
arr2where we will store our 'chunks'. - The for loop starts at zero, increments by
sizeeach time through the loop, and stops when it reachesarr.length. - Note that this for loop does not loop through
arr. Instead, we are using the loop to generate numbers we can use as indices to slice the array in the right locations. - Inside our loop, we create each chunk using
arr.slice(i, i+size), and add this value toarr2witharr2.push(). - Finally, we return the value of
arr2.
Relevant Links
Solution 3 (Click to Show/Hide)
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);
Code Explanation
-
Firstly, we create two variables.
newArris an empty array which we will push to. We also have theivariable set to zero, for use in our while loop. -
Our while loop loops until
iis equal to or more than the length of the array in our test. -
Inside our loop, we push to the
newArrarray usingarr.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 ofsizeontoi. -
Finally, we return the value of
newArr.
Relevant Links
Solution 4 (Click to Show/Hide)
function chunkArrayInGroups(arr, size) {
var newArr = [];
while (arr.length) {
newArr.push(arr.splice(0, size));
}
return newArr;
}
Code Explanation
- Firstly, we create a variable.
newArris an empty array which we will push to. - Our
whileloop loops until the length of the array in our test is not 0. - Inside our loop, we push to the
newArrarray usingarr.splice(0, size). - For each iteration of
whileloop, it deletessizenumber of elements from the front ofarrand push them as an array tonewArr. - Finally, we return the value of
newArr.
Relevant Links
Solution 5 (Click to Show/Hide)
function chunkArrayInGroups(arr, size) {
if (arr.length <= size) {
return [arr];
} else {
return [arr.slice(0, size)].concat(
chunkArrayInGroups(arr.slice(size), size)
);
}
}
Code Explanation
- Array smaller than size is returned nested.
- For any array larger than size, it is split in two. First segment is nested and concatenated with second segment which makes a recursive call.