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Commit c1cf57a

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fix(book/set): split Set chapter into Hash Set and Tree Set for better
coverage
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-351
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‎README.md

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#### [HashMaps](book/content/part03/map.asc)
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Learn how to implement different types of Maps such as:
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- [HashMap](book/content/part03/hashmap.asc)
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- [TreeMap](book/content/part03/treemap.asc)
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- [HashMap](book/content/part02/hash-map.asc)
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- [TreeMap](book/content/part03/tree-map.asc)
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Also, [learn the difference between the different Maps implementations](book/content/part03/time-complexity-graph-data-structures.asc):
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‎book/B-self-balancing-binary-search-trees.asc

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Right rotation moves a node on the right as a child of another node.
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Take a look at the `@example` in the code below.
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As you can see we have an unbalanced tree `4-3-2-1`.
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Take a look at the examples in the code in the next section.
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As you will see we have an unbalanced tree `4-3-2-1`.
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We want to balance the tree, for that we need to do a right rotation of node 3.
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So, we move node 3 as the right child of the previous child.
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=== Self-balancing trees implementations
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So far, we have study how to make tree rotations which are the basis for self-balancing trees. There are different implementations of self-balancing trees such a Red-Black Tree and AVL Tree.
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‎book/D-interview-questions-solutions.asc

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[#hashmap-q-two-sum]
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include::content/part02/hash-map.asc[tag=hashmap-q-two-sum]
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// include::content/part03/hashmap.asc[tag=hashmap-q-two-sum]
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// include::content/part02/hash-map.asc[tag=hashmap-q-two-sum]
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This simple problem can have many solutions; let's explore some.
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[#hashmap-q-subarray-sum-equals-k]
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include::content/part02/hash-map.asc[tag=hashmap-q-subarray-sum-equals-k]
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// include::content/part03/hashmap.asc[tag=hashmap-q-subarray-sum-equals-k]
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// include::content/part02/hash-map.asc[tag=hashmap-q-subarray-sum-equals-k]
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This problem has multiple ways to solve it. Let's explore some.
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[#set-q-most-common-word]
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include::content/part03/set.asc[tag=set-q-most-common-word]
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include::content/part02/hash-set.asc[tag=set-q-most-common-word]
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This problem requires multiple steps. We can use a `Set` for quickly looking up banned words. For getting the count of each word, we used a `Map`.
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[#set-q-longest-substring-without-repeating-characters]
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include::content/part03/set.asc[tag=set-q-longest-substring-without-repeating-characters]
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include::content/part02/hash-set.asc[tag=set-q-longest-substring-without-repeating-characters]
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One of the most efficient ways to find repeating characters is using a `Map` or `Set`. Use a `Map` when you need to keep track of the count/index (e.g., string -> count) and use a `Set` when you only need to know if there are repeated characters or not.
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‎book/content/part02/hash-map.asc

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(((Map))) (((HashMap))) (((HashTable))) (((Data Structures, Linear, HashMap)))
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[[hashmap-chap]]
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=== Hash Map
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=== Map
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A Map is a data structure where a `key` is mapped to a `value`. It's used for a fast lookup of values based on the given key. Only one key can map to a value (no duplicates).
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JavaScript has two ways to use Maps: one uses objects (`{}`), and the other is using the built-in `Map`.
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[[hashmap-examples]]
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.Using Objects as a HashMap.
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[source, javascript]
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----

‎book/content/part02/hash-set.asc

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ifndef::imagesdir[]
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:imagesdir: ../../images
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:codedir: ../../../src
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endif::[]
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(((Set))) (((Data Structures, Non-Linear, Set)))
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[[hash-set-chap]]
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=== Set
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Set is a data structure that allows you to store unique values. If you try to add the same value, multiple times only one instance will be added. Also, you can check very quickly if a value exists or not. Searching by value on arrays takes `O(n)`. However, searching by value on a Set takes `O(1)` on average.
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A Set can be implemented on different ways. One way it's using a <<hashmap-chap, Hash Map>> and other is using a <<tree-map-chap, Tree Map>>. JavaScript has a built-in Hash Set, so that' the one we are going to focus on.
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TIP: We will go more in details with <<tree-map-chap, Tree Map>> after we cover the <<binary-search-tree-chap>>.
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==== Set vs Array
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An array allows you to search a value by index in constant time `O(1)`, however if you don't know the index, searching a value would take you linear time `O(n)`. A Set has doesn't allow you to search value by index, but you can search by value in constant time. The `Set.add` and `Set.has` method both are `O(1)` in average.
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Take a look at the following examples:
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.Set usage example (using JavaScript built-in Set)
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[source, javascript]
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----
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const set = new Set();
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set.add(1); //↪️ Set [ 1 ]
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set.add(1); //↪️ Set [ 1 ]
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set.add(2); //↪️ Set [ 1, 2 ]
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set.add(3); //↪️ Set [ 1, 2, 3 ]
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set.has(1); //↪️ true
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set.delete(1); //↪️ removes 1 from the set
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set.has(1); //↪️ false, 1 has been removed
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set.size; //↪️ 2, we just removed one value
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console.log(set); //↪️ Set(2) {2, 3}
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----
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As you can see, even if we insert the same value multiple times, it only gets added once.
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Similar to a <<hashmap-examples, map>>, you can also insert objects and any kind of objects. However, be careful, because anything that is not a number, string or symbol would be matched by reference. Let's do some examples.
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.Using a Set with objects
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[source, javascript]
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----
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const set = new Set();
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set.add({a: 1, b: 2});
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set.has({a: 1, b: 2}); // ↪️ false
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const a = {a: 1, b: 2};
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set.add(a);
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set.has(a); // ↪️ true
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console.log(set); // Set { [ 1, 2, 3 ], [ 1, 2, 3 ] }
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----
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As you can see, you can't to find object using a new object (e.g. `{a: 1, b: 2}`), you need the reference to find it.
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If you need to match by value, you would need to convert it to an string using `JSON.stringify`.
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.Workaround to find objects by value.
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[source, javascript]
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----
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const set = new Set();
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set.add(JSON.stringify({a: 1, b: 2}));
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set.has(JSON.stringify({a: 1, b: 2})); // ↪️ true
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console.log(set); // Set { '{"a":1,"b":2}' }
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----
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==== Removing duplicates from an array.
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One common case for a Set is to eliminate duplicates from an array.
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.Removing duplicates from an array
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[source, javascript]
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----
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const arr = [1, 2, 2, 1, 3, 2];
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// convert array to set
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const set = new Set(arr);
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// convert set to array
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const uniqueValues = Array.from(set);
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// check array
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console.log(uniqueValues); // [ 1, 2, 3 ]
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----
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You can also do it all in one line.
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.One-liner to remove duplicates from array.
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[source, javascript]
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----
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const arr = [1, 2, 2, 1, 3, 2];
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console.log([...new Set(arr)]); // [ 1, 2, 3 ]
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----
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==== Time Complexity of a Hash Set
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All operation on Hash Set are constant time on average: `O(1)`. Similar to the Hash Map, there are cases when the the Set is getting full and it would do a rehash taking `O(n)` for that one insertion.
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// tag::table[]
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.Time complexity HashSet
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|===
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.2+.^s| Data Structure 2+^s| Searching By .2+^.^s| Insert .2+^.^s| Delete .2+^.^s| Space Complexity
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^|_Index/Key_ ^|_Value_
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| Hash Set ^|O(1) ^|- ^|O(1)* ^|O(1) ^|O(n)
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|===
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{empty}* = Amortized run time. E.g. rehashing might affect run time to *O(n)*.
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// end::table[]
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==== Practice Questions
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(((Interview Questions, Set)))
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// tag::set-q-most-common-word[]
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===== Most common word
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*ST-1*) _Given a text and a list of banned words.
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Find the most common word that is not on the banned list.
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You might need to sanitize the text and strip out punctuation `?!,'.`_
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// end::set-q-most-common-word[]
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// _Seen in interviews at: Amazon._
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Examples:
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[source, javascript]
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----
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mostCommonWord(
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`How much wood, would a Woodchuck chuck,
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if a woodchuck could chuck?`,
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['a'],
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); // woodchuck or chuck (both show up twice)
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mostCommonWord(
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`It's a blue ball and its shade... Very BLUE!`,
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['and']); // blue (it show up twice, "it" and "its" once)
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----
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Starter code:
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[source, javascript]
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----
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include::../../interview-questions/most-common-word.js[tags=description;placeholder]
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----
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_Solution: <<set-q-most-common-word>>_
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// tag::set-q-longest-substring-without-repeating-characters[]
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===== Longest Without Repeating
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*ST-2*) _Find the length of the longest substring without repeating characters._
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// end::set-q-longest-substring-without-repeating-characters[]
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// _Seen in interviews at: Amazon, Facebook, Bloomberg._
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Examples:
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[source, javascript]
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----
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lenLongestSubstring('aaaaa'); // 1 ('a')
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lenLongestSubstring('abccdefg'); // 5 ('cdefg')
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lenLongestSubstring('abc'); // 3 ('abc')
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----
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Starter code:
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[source, javascript]
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----
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include::../../interview-questions/longest-substring-without-repeating-characters.js[tags=description;placeholder]
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----
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_Solution: <<set-q-longest-substring-without-repeating-characters>>_

‎book/content/part03/binary-search-tree.asc

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:codedir: ../../../src
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endif::[]
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=== Binary Search Tree
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(((Binary Search Tree)))
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(((BST)))
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(((Data Structures, Non-Linear, Binary Search Tree)))
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[[binary-search-tree-chap]]
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=== Binary Search Tree
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.To recap, the Binary Search Tree (BST) is a tree data structure that keeps the following constraints:
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* Each node must have at most two children. Usually referred to as "left" and "right".

‎book/content/part03/graph-search.asc

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c0, c1, c2 [color=midnightblue]
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// c3 [color=red]
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}
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a0, b0, c0 [label = 0]
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a1, b1, c1 [label = 1]
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a2, b2, c2 [label = 2]
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b3, c3 [label = 3]
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}
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....
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‎book/content/part03/graph.asc

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If the node doesn't exist, then we create the new node and add it to a `HashMap`.
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TIP: <<part03-graph-data-structures#map>> stores key/pair value very efficiently. Lookup is `O(1)`.
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TIP: <<tree-map-chap>> stores key/pair value very efficiently. Lookup is `O(1)`.
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The `key` is the node's value, while the `value` is the newly created node.
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