diff --git a/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/-iterate-through-the-keys-of-an-object-with-a-for...in-statement.english.md b/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/-iterate-through-the-keys-of-an-object-with-a-for...in-statement.english.md
index 6808afd4cc8..d144e693dbe 100644
--- a/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/-iterate-through-the-keys-of-an-object-with-a-for...in-statement.english.md
+++ b/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/-iterate-through-the-keys-of-an-object-with-a-for...in-statement.english.md
@@ -6,27 +6,20 @@ challengeType: 1
## Description
-Sometimes you may need to iterate through all the keys within an object. This requires a specific syntax in JavaScript called a for...in statement. In the example below, there is a users object and the loop iterates through it to display the user's name to the console.
-
-const users = {
- Alan: {
- age: 30
- },
- Jeff: {
- age: 45
- },
- Sarah: {
- age: 18
- }
- Ryan: {
- age: 24
- }
-};
-for (let user in users) {
- console.log(user);
-}
-// logs:
Alan
Jeff
Sarah
Ryan
-
+Sometimes you may need to iterate through all the keys within an object. This requires a specific syntax in JavaScript called a for...in statement. For our users object, this could look like:
+
+```js
+for (let user in users) {
+ console.log(user);
+}
+
+// logs:
+Alan
+Jeff
+Sarah
+Ryan
+```
+
In this statement, we defined a variable user, and as you can see, this variable was reset during each iteration to each of the object's keys as the statement looped through the object, resulting in each user's name being printed to the console.
NOTE: Objects do not maintain an ordering to stored keys like arrays do; thus a key's position on an object, or the relative order in which it appears, is irrelevant when referencing or accessing that key.
diff --git a/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/access-an-arrays-contents-using-bracket-notation.english.md b/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/access-an-arrays-contents-using-bracket-notation.english.md
index b6a16e29404..44ba88858f3 100644
--- a/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/access-an-arrays-contents-using-bracket-notation.english.md
+++ b/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/access-an-arrays-contents-using-bracket-notation.english.md
@@ -8,13 +8,27 @@ challengeType: 1
The fundamental feature of any data structure is, of course, the ability to not only store data, but to be able to retrieve that data on command. So, now that we've learned how to create an array, let's begin to think about how we can access that array's information.
When we define a simple array as seen below, there are 3 items in it:
-let ourArray = ["a", "b", "c"];
+
+```
+let ourArray = ["a", "b", "c"];
+```
+
In an array, each array item has an index. This index doubles as the position of that item in the array, and how you reference it. However, it is important to note, that JavaScript arrays are zero-indexed, meaning that the first element of an array is actually at the zeroth position, not the first.
In order to retrieve an element from an array we can enclose an index in brackets and append it to the end of an array, or more commonly, to a variable which references an array object. This is known as bracket notation.
For example, if we want to retrieve the "a" from ourArray and assign it to a variable, we can do so with the following code:
-let ourVariable = ourArray[0];
// ourVariable equals "a"
+
+```
+let ourVariable = ourArray[0];
+// ourVariable equals "a"
+```
+
In addition to accessing the value associated with an index, you can also set an index to a value using the same notation:
-ourArray[1] = "not b anymore";
// ourArray now equals ["a", "not b anymore", "c"];
+
+```
+ourArray[1] = "not b anymore";
+// ourArray now equals ["a", "not b anymore", "c"];
+```
+
Using bracket notation, we have now reset the item at index 1 from "b", to "not b anymore".
diff --git a/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/access-property-names-with-bracket-notation.english.md b/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/access-property-names-with-bracket-notation.english.md
index e2f59fb6ff8..7fe6b881bac 100644
--- a/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/access-property-names-with-bracket-notation.english.md
+++ b/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/access-property-names-with-bracket-notation.english.md
@@ -7,7 +7,12 @@ challengeType: 1
## Description
In the first object challenge we mentioned the use of bracket notation as a way to access property values using the evaluation of a variable. For instance, imagine that our foods object is being used in a program for a supermarket cash register. We have some function that sets the selectedFood and we want to check our foods object for the presence of that food. This might look like:
-let selectedFood = getCurrentFood(scannedItem);
let inventory = foods[selectedFood];
+
+```
+let selectedFood = getCurrentFood(scannedItem);
+let inventory = foods[selectedFood];
+```
+
This code will evaluate the value stored in the selectedFood variable and return the value of that key in the foods object, or undefined if it is not present. Bracket notation is very useful because sometimes object properties are not known before runtime or we need to access them in a more dynamic way.
diff --git a/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/add-items-to-an-array-with-push-and-unshift.english.md b/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/add-items-to-an-array-with-push-and-unshift.english.md
index d8b57dc09e2..64a967f62f9 100644
--- a/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/add-items-to-an-array-with-push-and-unshift.english.md
+++ b/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/add-items-to-an-array-with-push-and-unshift.english.md
@@ -8,8 +8,18 @@ challengeType: 1
An array's length, like the data types it can contain, is not fixed. Arrays can be defined with a length of any number of elements, and elements can be added or removed over time; in other words, arrays are mutable. In this challenge, we will look at two methods with which we can programmatically modify an array: Array.push() and Array.unshift().
Both methods take one or more elements as parameters and add those elements to the array the method is being called on; the push() method adds elements to the end of an array, and unshift() adds elements to the beginning. Consider the following:
-let twentyThree = 'XXIII';
let romanNumerals = ['XXI', 'XXII'];
romanNumerals.unshift('XIX', 'XX');
// now equals ['XIX', 'XX', 'XXI', 'XXII']
romanNumerals.push(twentyThree);
// now equals ['XIX', 'XX', 'XXI', 'XXII', 'XXIII']
-Notice that we can also pass variables, which allows us even greater flexibility in dynamically modifying our array's data.
+
+```
+let twentyThree = 'XXIII';
+let romanNumerals = ['XXI', 'XXII'];
+
+romanNumerals.unshift('XIX', 'XX');
+// now equals ['XIX', 'XX', 'XXI', 'XXII']
+
+romanNumerals.push(twentyThree);
+// now equals ['XIX', 'XX', 'XXI', 'XXII', 'XXIII']Notice that we can also pass variables, which allows us even greater flexibility in dynamically modifying our array's data.
+```
+
## Instructions
diff --git a/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/add-items-using-splice.english.md b/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/add-items-using-splice.english.md
index a422733d0db..09618f8b560 100644
--- a/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/add-items-using-splice.english.md
+++ b/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/add-items-using-splice.english.md
@@ -7,7 +7,20 @@ challengeType: 1
## Description
Remember in the last challenge we mentioned that splice() can take up to three parameters? Well, we can go one step further with splice() — in addition to removing elements, we can use that third parameter, which represents one or more elements, to add them as well. This can be incredibly useful for quickly switching out an element, or a set of elements, for another. For instance, let's say you're storing a color scheme for a set of DOM elements in an array, and want to dynamically change a color based on some action:
-function colorChange(arr, index, newColor) {
arr.splice(index, 1, newColor);
return arr;
}
let colorScheme = ['#878787', '#a08794', '#bb7e8c', '#c9b6be', '#d1becf'];
colorScheme = colorChange(colorScheme, 2, '#332327');
// we have removed '#bb7e8c' and added '#332327' in its place
// colorScheme now equals ['#878787', '#a08794', '#332327', '#c9b6be', '#d1becf']
+
+```
+function colorChange(arr, index, newColor) {
+ arr.splice(index, 1, newColor);
+ return arr;
+}
+
+let colorScheme = ['#878787', '#a08794', '#bb7e8c', '#c9b6be', '#d1becf'];
+
+colorScheme = colorChange(colorScheme, 2, '#332327');
+// we have removed '#bb7e8c' and added '#332327' in its place
+// colorScheme now equals ['#878787', '#a08794', '#332327', '#c9b6be', '#d1becf']
+```
+
This function takes an array of hex values, an index at which to remove an element, and the new color to replace the removed element with. The return value is an array containing a newly modified color scheme! While this example is a bit oversimplified, we can see the value that utilizing splice() to its maximum potential can have.
diff --git a/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/add-key-value-pairs-to-javascript-objects.english.md b/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/add-key-value-pairs-to-javascript-objects.english.md
index 0b05f01d74b..82f4e8f3acb 100644
--- a/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/add-key-value-pairs-to-javascript-objects.english.md
+++ b/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/add-key-value-pairs-to-javascript-objects.english.md
@@ -7,11 +7,30 @@ challengeType: 1
## Description
At their most basic, objects are just collections of key-value pairs, or in other words, pieces of data mapped to unique identifiers that we call properties or keys. Let's take a look at a very simple example:
-let FCC_User = {
username: 'awesome_coder',
followers: 572,
points: 1741,
completedProjects: 15
};
+
+```
+let FCC_User = {
+ username: 'awesome_coder',
+ followers: 572,
+ points: 1741,
+ completedProjects: 15
+};
+```
+
The above code defines an object called FCC_User that has four properties, each of which map to a specific value. If we wanted to know the number of followers FCC_User has, we can access that property by writing:
-let userData = FCC_User.followers;
// userData equals 572
+
+```
+let userData = FCC_User.followers;
+// userData equals 572
+```
+
This is called dot notation. Alternatively, we can also access the property with brackets, like so:
-let userData = FCC_User['followers']
// userData equals 572
+
+```
+let userData = FCC_User['followers']
+// userData equals 572
+```
+
Notice that with bracket notation, we enclosed followers in quotes. This is because the brackets actually allow us to pass a variable in to be evaluated as a property name (hint: keep this in mind for later!). Had we passed followers in without the quotes, the JavaScript engine would have attempted to evaluate it as a variable, and a ReferenceError: followers is not defined would have been thrown.
diff --git a/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/check-for-the-presence-of-an-element-with-indexof.english.md b/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/check-for-the-presence-of-an-element-with-indexof.english.md
index 6e264df7613..3526491596b 100644
--- a/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/check-for-the-presence-of-an-element-with-indexof.english.md
+++ b/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/check-for-the-presence-of-an-element-with-indexof.english.md
@@ -8,7 +8,15 @@ challengeType: 1
Since arrays can be changed, or mutated, at any time, there's no guarantee about where a particular piece of data will be on a given array, or if that element even still exists. Luckily, JavaScript provides us with another built-in method, indexOf(), that allows us to quickly and easily check for the presence of an element on an array. indexOf() takes an element as a parameter, and when called, it returns the position, or index, of that element, or -1 if the element does not exist on the array.
For example:
-let fruits = ['apples', 'pears', 'oranges', 'peaches', 'pears'];
fruits.indexOf('dates') // returns -1
fruits.indexOf('oranges') // returns 2
fruits.indexOf('pears') // returns 1, the first index at which the element exists
+
+```
+let fruits = ['apples', 'pears', 'oranges', 'peaches', 'pears'];
+
+fruits.indexOf('dates') // returns -1
+fruits.indexOf('oranges') // returns 2
+fruits.indexOf('pears') // returns 1, the first index at which the element exists
+```
+
## Instructions
diff --git a/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/check-if-an-object-has-a-property.english.md b/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/check-if-an-object-has-a-property.english.md
index 8deb17c5693..2c2eb6fc223 100644
--- a/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/check-if-an-object-has-a-property.english.md
+++ b/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/check-if-an-object-has-a-property.english.md
@@ -7,7 +7,13 @@ challengeType: 1
## Description
Now we can add, modify, and remove keys from objects. But what if we just wanted to know if an object has a specific property? JavaScript provides us with two different ways to do this. One uses the hasOwnProperty() method and the other uses the in keyword. If we have an object users with a property of Alan, we could check for its presence in either of the following ways:
-users.hasOwnProperty('Alan');
'Alan' in users;
// both return true
+
+```
+users.hasOwnProperty('Alan');
+'Alan' in users;
+// both return true
+```
+
## Instructions
diff --git a/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/combine-arrays-with-the-spread-operator.english.md b/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/combine-arrays-with-the-spread-operator.english.md
index 2302651eba8..94379bfa1ac 100644
--- a/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/combine-arrays-with-the-spread-operator.english.md
+++ b/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/combine-arrays-with-the-spread-operator.english.md
@@ -7,7 +7,14 @@ challengeType: 1
## Description
Another huge advantage of the spread operator, is the ability to combine arrays, or to insert all the elements of one array into another, at any index. With more traditional syntaxes, we can concatenate arrays, but this only allows us to combine arrays at the end of one, and at the start of another. Spread syntax makes the following operation extremely simple:
-let thisArray = ['sage', 'rosemary', 'parsley', 'thyme'];
let thatArray = ['basil', 'cilantro', ...thisArray, 'coriander'];
// thatArray now equals ['basil', 'cilantro', 'sage', 'rosemary', 'parsley', 'thyme', 'coriander']
+
+```
+let thisArray = ['sage', 'rosemary', 'parsley', 'thyme'];
+
+let thatArray = ['basil', 'cilantro', ...thisArray, 'coriander'];
+// thatArray now equals ['basil', 'cilantro', 'sage', 'rosemary', 'parsley', 'thyme', 'coriander']
+```
+
Using spread syntax, we have just achieved an operation that would have been more complex and more verbose had we used traditional methods.
diff --git a/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/copy-an-array-with-the-spread-operator.english.md b/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/copy-an-array-with-the-spread-operator.english.md
index 4dc22dca24c..820d0cb1358 100644
--- a/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/copy-an-array-with-the-spread-operator.english.md
+++ b/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/copy-an-array-with-the-spread-operator.english.md
@@ -8,7 +8,14 @@ challengeType: 1
While slice() allows us to be selective about what elements of an array to copy, among several other useful tasks, ES6's new spread operator allows us to easily copy all of an array's elements, in order, with a simple and highly readable syntax. The spread syntax simply looks like this: ...
In practice, we can use the spread operator to copy an array like so:
-let thisArray = [true, true, undefined, false, null];
let thatArray = [...thisArray];
// thatArray equals [true, true, undefined, false, null]
// thisArray remains unchanged, and is identical to thatArray
+
+```
+let thisArray = [true, true, undefined, false, null];
+let thatArray = [...thisArray];
+// thatArray equals [true, true, undefined, false, null]
+// thisArray remains unchanged, and is identical to thatArray
+```
+
## Instructions
diff --git a/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/copy-array-items-using-slice.english.md b/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/copy-array-items-using-slice.english.md
index ffe3d3d2412..4383d95faea 100644
--- a/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/copy-array-items-using-slice.english.md
+++ b/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/copy-array-items-using-slice.english.md
@@ -7,7 +7,16 @@ challengeType: 1
## Description
The next method we will cover is slice(). slice(), rather than modifying an array, copies, or extracts, a given number of elements to a new array, leaving the array it is called upon untouched. slice() takes only 2 parameters — the first is the index at which to begin extraction, and the second is the index at which to stop extraction (extraction will occur up to, but not including the element at this index). Consider this:
-let weatherConditions = ['rain', 'snow', 'sleet', 'hail', 'clear'];
let todaysWeather = weatherConditions.slice(1, 3);
// todaysWeather equals ['snow', 'sleet'];
// weatherConditions still equals ['rain', 'snow', 'sleet', 'hail', 'clear']
+
+```
+let weatherConditions = ['rain', 'snow', 'sleet', 'hail', 'clear'];
+
+let todaysWeather = weatherConditions.slice(1, 3);
+// todaysWeather equals ['snow', 'sleet'];
+// weatherConditions still equals ['rain', 'snow', 'sleet', 'hail', 'clear']
+
+```
+
In effect, we have created a new array by extracting elements from an existing array.
diff --git a/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/create-complex-multi-dimensional-arrays.english.md b/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/create-complex-multi-dimensional-arrays.english.md
index 4e34163f7e9..263a3dc92ed 100644
--- a/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/create-complex-multi-dimensional-arrays.english.md
+++ b/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/create-complex-multi-dimensional-arrays.english.md
@@ -8,12 +8,43 @@ challengeType: 1
Awesome! You have just learned a ton about arrays! This has been a fairly high level overview, and there is plenty more to learn about working with arrays, much of which you will see in later sections. But before moving on to looking at Objects, lets take one more look, and see how arrays can become a bit more complex than what we have seen in previous challenges.
One of the most powerful features when thinking of arrays as data structures, is that arrays can contain, or even be completely made up of other arrays. We have seen arrays that contain arrays in previous challenges, but fairly simple ones. However, arrays can contain an infinite depth of arrays that can contain other arrays, each with their own arbitrary levels of depth, and so on. In this way, an array can very quickly become very complex data structure, known as a multi-dimensional, or nested array. Consider the following example:
-let nestedArray = [ // top, or first level - the outer most array
['deep'], // an array within an array, 2 levels of depth
[
['deeper'], ['deeper'] // 2 arrays nested 3 levels deep
],
[
[
['deepest'], ['deepest'] // 2 arrays nested 4 levels deep
],
[
[
['deepest-est?'] // an array nested 5 levels deep
]
]
]
];
+
+```
+let nestedArray = [ // top, or first level - the outer most array
+ ['deep'], // an array within an array, 2 levels of depth
+ [
+ ['deeper'], ['deeper'] // 2 arrays nested 3 levels deep
+ ],
+ [
+ [
+ ['deepest'], ['deepest'] // 2 arrays nested 4 levels deep
+ ],
+ [
+ [
+ ['deepest-est?'] // an array nested 5 levels deep
+ ]
+ ]
+ ]
+];
+```
+
While this example may seem convoluted, this level of complexity is not unheard of, or even unusual, when dealing with large amounts of data.
However, we can still very easily access the deepest levels of an array this complex with bracket notation:
-console.log(nestedArray[2][1][0][0][0]);
// logs: deepest-est?
+
+```
+console.log(nestedArray[2][1][0][0][0]);
+// logs: deepest-est?
+```
+
And now that we know where that piece of data is, we can reset it if we need to:
-nestedArray[2][1][0][0][0] = 'deeper still';
console.log(nestedArray[2][1][0][0][0]);
// now logs: deeper still
+
+```
+nestedArray[2][1][0][0][0] = 'deeper still';
+
+console.log(nestedArray[2][1][0][0][0]);
+// now logs: deeper still
+```
+
## Instructions
diff --git a/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/iterate-through-all-an-arrays-items-using-for-loops.english.md b/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/iterate-through-all-an-arrays-items-using-for-loops.english.md
index 47be6a7ff23..032734432c3 100644
--- a/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/iterate-through-all-an-arrays-items-using-for-loops.english.md
+++ b/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/iterate-through-all-an-arrays-items-using-for-loops.english.md
@@ -8,7 +8,22 @@ challengeType: 1
Sometimes when working with arrays, it is very handy to be able to iterate through each item to find one or more elements that we might need, or to manipulate an array based on which data items meet a certain set of criteria. JavaScript offers several built in methods that each iterate over arrays in slightly different ways to achieve different results (such as every(), forEach(), map(), etc.), however the technique which is most flexible and offers us the greatest amount of control is a simple for loop.
Consider the following:
-function greaterThanTen(arr) {
let newArr = [];
for (let i = 0; i < arr.length; i++) {
if (arr[i] > 10) {
newArr.push(arr[i]);
}
}
return newArr;
}
greaterThanTen([2, 12, 8, 14, 80, 0, 1]);
// returns [12, 14, 80]
+
+```
+function greaterThanTen(arr) {
+ let newArr = [];
+ for (let i = 0; i < arr.length; i++) {
+ if (arr[i] > 10) {
+ newArr.push(arr[i]);
+ }
+ }
+ return newArr;
+}
+
+greaterThanTen([2, 12, 8, 14, 80, 0, 1]);
+// returns [12, 14, 80]
+```
+
Using a for loop, this function iterates through and accesses each element of the array, and subjects it to a simple test that we have created. In this way, we have easily and programmatically determined which data items are greater than 10, and returned a new array containing those items.
diff --git a/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/modify-an-object-nested-within-an-object.english.md b/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/modify-an-object-nested-within-an-object.english.md
index bcbe0115aed..dde94c73902 100644
--- a/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/modify-an-object-nested-within-an-object.english.md
+++ b/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/modify-an-object-nested-within-an-object.english.md
@@ -7,7 +7,22 @@ challengeType: 1
## Description
Now let's take a look at a slightly more complex object. Object properties can be nested to an arbitrary depth, and their values can be any type of data supported by JavaScript, including arrays and even other objects. Consider the following:
-let nestedObject = {
id: 28802695164,
date: 'December 31, 2016',
data: {
totalUsers: 99,
online: 80,
onlineStatus: {
active: 67,
away: 13
}
}
};
+
+```
+let nestedObject = {
+ id: 28802695164,
+ date: 'December 31, 2016',
+ data: {
+ totalUsers: 99,
+ online: 80,
+ onlineStatus: {
+ active: 67,
+ away: 13
+ }
+ }
+};
+```
+
nestedObject has three unique keys: id, whose value is a number, date whose value is a string, and data, whose value is an object which has yet another object nested within it. While structures can quickly become complex, we can still use the same notations to access the information we need.
diff --git a/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/remove-items-from-an-array-with-pop-and-shift.english.md b/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/remove-items-from-an-array-with-pop-and-shift.english.md
index 22355f3a807..e732d3ff48e 100644
--- a/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/remove-items-from-an-array-with-pop-and-shift.english.md
+++ b/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/remove-items-from-an-array-with-pop-and-shift.english.md
@@ -8,9 +8,25 @@ challengeType: 1
Both push() and unshift() have corresponding methods that are nearly functional opposites: pop() and shift(). As you may have guessed by now, instead of adding, pop() removes an element from the end of an array, while shift() removes an element from the beginning. The key difference between pop() and shift() and their cousins push() and unshift(), is that neither method takes parameters, and each only allows an array to be modified by a single element at a time.
Let's take a look:
-let greetings = ['whats up?', 'hello', 'see ya!'];
greetings.pop();
// now equals ['whats up?', 'hello']
greetings.shift();
// now equals ['hello']
+
+```
+let greetings = ['whats up?', 'hello', 'see ya!'];
+
+greetings.pop();
+// now equals ['whats up?', 'hello']
+
+greetings.shift();
+// now equals ['hello']
+```
+
We can also return the value of the removed element with either method like this:
-let popped = greetings.pop();
// returns 'hello'
// greetings now equals []
+
+```
+let popped = greetings.pop();
+// returns 'hello'
+// greetings now equals []
+```
+
## Instructions
diff --git a/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/remove-items-using-splice.english.md b/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/remove-items-using-splice.english.md
index c342af6b4d1..1f62a67435b 100644
--- a/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/remove-items-using-splice.english.md
+++ b/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/remove-items-using-splice.english.md
@@ -8,9 +8,24 @@ challengeType: 1
Ok, so we've learned how to remove elements from the beginning and end of arrays using shift() and pop(), but what if we want to remove an element from somewhere in the middle? Or remove more than one element at once? Well, that's where splice() comes in. splice() allows us to do just that: remove any number of consecutive elements from anywhere in an array.
splice() can take up to 3 parameters, but for now, we'll focus on just the first 2. The first two parameters of splice() are integers which represent indexes, or positions, of the array that splice() is being called upon. And remember, arrays are zero-indexed, so to indicate the first element of an array, we would use 0. splice()'s first parameter represents the index on the array from which to begin removing elements, while the second parameter indicates the number of elements to delete. For example:
-let array = ['today', 'was', 'not', 'so', 'great'];
array.splice(2, 2);
// remove 2 elements beginning with the 3rd element
// array now equals ['today', 'was', 'great']
+
+```
+let array = ['today', 'was', 'not', 'so', 'great'];
+
+array.splice(2, 2);
+// remove 2 elements beginning with the 3rd element
+// array now equals ['today', 'was', 'great']
+```
+
splice() not only modifies the array it's being called on, but it also returns a new array containing the value of the removed elements:
-let array = ['I', 'am', 'feeling', 'really', 'happy'];
let newArray = array.splice(3, 2);
// newArray equals ['really', 'happy']
+
+```
+let array = ['I', 'am', 'feeling', 'really', 'happy'];
+
+let newArray = array.splice(3, 2);
+// newArray equals ['really', 'happy']
+```
+
## Instructions
diff --git a/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/use-an-array-to-store-a-collection-of-data.english.md b/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/use-an-array-to-store-a-collection-of-data.english.md
index 0a810808c55..d52d46bce93 100644
--- a/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/use-an-array-to-store-a-collection-of-data.english.md
+++ b/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/use-an-array-to-store-a-collection-of-data.english.md
@@ -7,10 +7,41 @@ challengeType: 1
## Description
The below is an example of the simplest implementation of an array data structure. This is known as a one-dimensional array, meaning it only has one level, or that it does not have any other arrays nested within it. Notice it contains booleans, strings, and numbers, among other valid JavaScript data types:
-let simpleArray = ['one', 2, 'three’, true, false, undefined, null];
console.log(simpleArray.length);
// logs 7
+
+```
+let simpleArray = ['one', 2, 'three', true, false, undefined, null];
+console.log(simpleArray.length);
+// logs 7
+```
+
All arrays have a length property, which as shown above, can be very easily accessed with the syntax Array.length.
A more complex implementation of an array can be seen below. This is known as a multi-dimensional array, or an array that contains other arrays. Notice that this array also contains JavaScript objects, which we will examine very closely in our next section, but for now, all you need to know is that arrays are also capable of storing complex objects.
-let complexArray = [
[
{
one: 1,
two: 2
},
{
three: 3,
four: 4
}
],
[
{
a: "a",
b: "b"
},
{
c: "c",
d: “d”
}
]
];
+
+```
+let complexArray = [
+ [
+ {
+ one: 1,
+ two: 2
+ },
+ {
+ three: 3,
+ four: 4
+ }
+ ],
+ [
+ {
+ a: "a",
+ b: "b"
+ },
+ {
+ c: "c",
+ d: "d"
+ }
+ ]
+];
+```
+
## Instructions
diff --git a/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/use-the-delete-keyword-to-remove-object-properties.english.md b/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/use-the-delete-keyword-to-remove-object-properties.english.md
index cf151a65da4..6ffa343a09f 100644
--- a/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/use-the-delete-keyword-to-remove-object-properties.english.md
+++ b/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-data-structures/use-the-delete-keyword-to-remove-object-properties.english.md
@@ -9,7 +9,11 @@ challengeType: 1
Now you know what objects are and their basic features and advantages. In short, they are key-value stores which provide a flexible, intuitive way to structure data, and, they provide very fast lookup time. Throughout the rest of these challenges, we will describe several common operations you can perform on objects so you can become comfortable applying these useful data structures in your programs.
In earlier challenges, we have both added to and modified an object's key-value pairs. Here we will see how we can remove a key-value pair from an object.
Let's revisit our foods object example one last time. If we wanted to remove the apples key, we can remove it by using the delete keyword like this:
-delete foods.apples;
+
+```
+delete foods.apples;
+```
+
## Instructions