Merge pull request amejiarosario#75 from amejiarosario/feat/set-questions

feat(book/set): add questions and solutions

Author: amejiarosario

book/D-interview-questions-solutions.asc

@@ -577,35 +577,81 @@ The sum is 1, however `sum - k` is `0`. If it doesn't exist on the map, we will
 - Time: `O(n)`. We visit every number once.
 - Space: `O(n)`. The map size will be the same as the original array.
 
-// :leveloffset: +1
 
-// === Solutions for TOPIC Questions
-// (((Interview Questions Solutions, TOPIC)))
 
-// :leveloffset: -1
+:leveloffset: +1
+
+=== Solutions for Set Questions
+(((Interview Questions Solutions, Set)))
+
+:leveloffset: -1
+
+
+[#set-q-most-common-word]
+include::content/part03/set.asc[tag=set-q-most-common-word]
+
+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`.
+
+*Algorithm*:
+
+- Convert text to lowercase.
+- Remove any special characters `!?',;.`.
+- Convert the paragraph into words array.
+- Count how many times words occur.
+- Exclude banned words from counts.
+- Return the word (or first one) that is the most repeated.
+
+*Implementation*:
+
+[source, javascript]
+----
+include::interview-questions/most-common-word.js[tags=description;solution]
+----
+
+Here are heavily relying on Regular Expressions:
+
+- `\W+` would match all non-words.
+- `\s+` catches all whitespace.
+
+The line that is mapping words to count seems very busy. Here's another version of the same code a little bit more explicit:
+
+[source, javascript]
+----
+include::interview-questions/most-common-word.js[tags=explicit]
+----
+
+*Complexity Analysis*:
+
+- Time: `O(m + n)`, where `n` is paragraph length and `m` is the number of banned words. If we were NOT using a `Set` for prohibited words, then the runtime would have been `O(mn)`.
+- Space: `O(m + n)`. The extra space complexity is given by the size of the `Map` and `Set`.
+
 
-// [#TOPIC-q-FILENAME]
-// include::content/part03/TOPIC_FILE.asc[tag=TOPIC-q-FILENAME]
 
-// RESTATE REQUIREMENTS AND DESCRIPTIONS
 
-// *Algorithm*:
 
-// - STEP 1
-// - STEP 2
-//     - STEP 2.1
-//     - STEP 2.2
+[#set-q-longest-substring-without-repeating-characters]
+include::content/part03/set.asc[tag=set-q-longest-substring-without-repeating-characters]
 
-// *Implementation*:
+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.
 
-// [source, javascript]
-// ----
-// include::interview-questions/FILENAME.js[tags=description;solution]
-// ----
+*Algorithm*:
+
+* Visit each letter.
+** Insert the letter on a Set.
+** Keep track of the maximum size of the Set in `max`.
+** If the letter has been seen before, delete until there's no duplicate.
+* Return max.
 
-// IMPLEMENTATION NOTES
+*Implementation*:
 
-// *Complexity Analysis*:
+[source, javascript]
+----
+include::interview-questions/longest-substring-without-repeating-characters.js[tags=description;solution]
+----
+
+We could also have used a Map and keep track of the indexes, but that's not necessary. In this case, the `Set` is all we need.
+
+*Complexity Analysis*:
 
-// - Time: `O(?)`. WHY?
-// - Space: `O(?)`. WHY?
+- Time: `O(n)`. We visit each letter once.
+- Space: `O(W)`, where `W` is the max length of non-repeating characters. The maximum size of the Set gives the space complexity. In the worst-case scenario, all letters are unique (`W = n`), so our space complexity would be `O(n)`. In the avg. case where there are one or more duplicates, it uses less space than `n`, because `W < n`.

book/content/part03/set.asc

@@ -7,9 +7,9 @@ endif::[]
 === Set
 (((Set)))
 (((Data Structures, Non-Linear, Set)))
-A set is a data structure where duplicated entries are not allowed. Set is like an array with unique values.
+A set is a data structure where duplicated entries are not allowed. A Set is like an array with only unique values.
 
-NOTE: JavaScript has already a built-in Set data structure.
+NOTE: JavaScript already has a built-in Set data structure.
 
 Take a look at the following
 example:
@@ -38,15 +38,15 @@ TIP: A hint... it should perform all operations in *O(1)** or at most *O(log n)*
 
 If we use a `map`, we can accomplish this. However, maps use a key/value pair. If we only use the keys, we can avoid duplicates. Since in a `map` you can only have one key at a time.
 
-As you might remember from the <<part03-graph-data-structures#map>> chapter, there are two ways of implementing a `map` and both can be used to create a `set`.  Let's explore the difference between the two implementations are.
+As you might remember from the <<part03-graph-data-structures#map>> chapter, there are two ways of implementing a `map`, and both can be used to create a `set`.  Let's explore the difference between the two implementations are.
 
 ==== HashSet vs TreeSet
 
-We can implement a `map` using a *balanced BST* and using a *hash function*. If we use them to implement a `Set`, then we would have a `HashSet` and `TreeSet` respectively.
+We can implement a `map` using a *balanced BST* or a *hash function*. If we use them to implement a `Set`, we would have a `HashSet` and `TreeSet`.
 
 * `TreeSet`, would return the values sorted in ascending order.
 * `HashSet`, would return the values in insertion order.
-* Operations on a `HashSet` would take on average O(1) and in the worst case (rehash is due), it would take O(n).
+* Operations on a `HashSet` would take on average O(1), and in the worst case (rehash is due), it would take O(n).
 * Operation on a `TreeSet` is always O(log n).
 
 Let’s implement both!
@@ -65,7 +65,7 @@ include::{codedir}/data-structures/sets/tree-set.js[tag=constructor]
 ----
 <1> Converts an array or any iterable data structure to a set.
 
-A common use case for Sets is to remove duplicated values from an array. We can do that by passing them in the constructor as follows:
+An everyday use case for Sets is to remove duplicated values from an array. We can do that bypassing them in the constructor as follows:
 
 .Removing duplicates from an Array using a Set
 [source, javascript]
@@ -115,7 +115,7 @@ Voilà! That’s it!
 
 ===== Converting TreeSet to Array
 
-A common use case for a Set is to convert it to an array or use in an iterator (for loops, forEach, …). Let’s provide the method for that:
+Another use case for a Set is to convert it to an array or use an iterator (for loops, forEach, …). Let’s provide the method for that:
 
 .TreeSet's iterator
 [source, javascript]
@@ -151,7 +151,7 @@ No more duplicates in our array!
 
 Check out our https://github.com/amejiarosario/dsa.js/blob/f69b744a1bddd3d99243ca64b3ad46f3f2dd7342/src/data-structures/sets/tree-set.js#L12[GitHub repo for the full TreeSet implementation].
 
-Let’s now, implement a `HashSet`.
+Let’s now implement a `HashSet`.
 
 [[hashset]]
 ==== HashSet
@@ -172,7 +172,7 @@ This constructor is useful for converting an array to set and initializing the `
 
 ===== Inserting values to a HashSet
 
-To insert items in a HashSet we use the `set` method of the `HashMap`:
+To insert items in a HashSet, we use the `set` method of the `HashMap`:
 
 .HashSet's `add` method
 [source, javascript]
@@ -181,7 +181,7 @@ include::{codedir}/data-structures/sets/hash-set.js[tag=add, indent=0]
 }
 ----
 
-`HashMap` stores key/value pairs, but for this, we only need the key, and we ignore the value.
+`HashMap` stores key/value pairs, but we only need the keys for Set, so we ignore the value.
 
 ===== Finding values in a HashSet
 
@@ -198,7 +198,7 @@ true, and if it’s empty, it will be false.
 
 ===== Deleting values from a HashSet
 
-For deleting a value from a hashSet we use the HashMap’s delete method:
+For deleting a value from a hashSet, we use the HashMap’s delete method:
 
 .HashSet's `delete` method
 [source, javascript]
@@ -210,7 +210,7 @@ This method has an average runtime of *O(1)*.
 
 ==== HashSet vs HashMap Time Complexity
 
-We can say that `HashMap` in on average more performant O(1) vs. O(log n). However, if a
+We can say that `HashMap` in on average, more performant O(1) vs. O(log n). However, if a
 rehash happens, it will take *O(n)* instead of *O(1)*. A `TreeSet` is always *O(log n)*.
 
 (((Tables, Non-Linear DS, HashSet/TreeSet complexities)))
@@ -236,3 +236,80 @@ difference besides runtime is that:
 .TreeSet vs HashSet
 * HashSet keeps data in insertion order
 * TreeSet keeps data sorted in ascending order.
+
+
+==== Practice Questions
+(((Interview Questions, Set)))
+
+// tag::set-q-most-common-word[]
+===== Most common word
+
+*ST-1*) _Given a text and a list of banned words.
+Find the most common word that is not on the banned list.
+You might need to sanitize the text and strip out punctuation `?!,'.`_
+// end::set-q-most-common-word[]
+
+// _Seen in interviews at: Amazon._
+
+Examples:
+
+[source, javascript]
+----
+mostCommonWord(
+  `How much wood, would a Woodchuck chuck,
+  if a woodchuck could chuck?`,
+  ['a'],
+); // woodchuck or chuck (both show up twice)
+
+mostCommonWord(
+`It's a blue ball and its shade... Very BLUE!`,
+['and']); // blue (it show up twice, "it" and "its" once)
+----
+
+Starter code:
+
+[source, javascript]
+----
+include::../../interview-questions/most-common-word.js[tags=description;placeholder]
+----
+
+
+_Solution: <<set-q-most-common-word>>_
+
+
+
+
+
+
+
+
+
+
+
+// tag::set-q-longest-substring-without-repeating-characters[]
+===== Longest Without Repeating
+
+*ST-2*) _Find the length of the longest substring without repeating characters._
+
+// end::set-q-longest-substring-without-repeating-characters[]
+
+// _Seen in interviews at: Amazon, Facebook, Bloomberg._
+
+Examples:
+
+[source, javascript]
+----
+lenLongestSubstring('aaaaa'); // 1 ('a')
+lenLongestSubstring('abccdefg'); // 5 ('cdefg')
+lenLongestSubstring('abc'); // 3 ('abc')
+----
+
+Starter code:
+
+[source, javascript]
+----
+include::../../interview-questions/longest-substring-without-repeating-characters.js[tags=description;placeholder]
+----
+
+
+_Solution: <<set-q-longest-substring-without-repeating-characters>>_

book/interview-questions/binary-tree-right-side-view.js

@@ -4,7 +4,7 @@ const { Queue } = require('../../src/index');
 /**
  * Find the rightmost nodes by level.
  *
- * @example
+ * @example rightSideView(bt([1,2,3,4])); // [1, 3, 4]
  *     1      <- 1
  *   /   \
  *  2     3   <- 3

book/interview-questions/buy-sell-stock.js

@@ -6,6 +6,7 @@
  *    maxProfit([1, 2, 3]); // => 2
  *    maxProfit([3, 2, 1]); // => 0
  * @param {number[]} prices - Array with daily stock prices
+ * @returns {number} - Max profit
  */
 function maxProfit(prices) {
 // end::description[]

book/interview-questions/daily-temperatures.js

@@ -9,6 +9,7 @@
  *  dailyTemperatures([73, 69, 72, 76, 73]); // [3, 1, 1, 0, 0]
  *
  * @param {number[]} t - Daily temperatures
+ * @returns {number[]} - Array with count of days to warmer temp.
  */
 function dailyTemperatures(t) {
   // end::description[]

book/interview-questions/linkedlist-same-data.js

@@ -9,8 +9,9 @@
  *    hasSameData(['he', 'll', 'o'], ['hel', 'lo']); // true
  *    hasSameData(['hel', 'lo'], ['hi']); // false
  *
- * @param {ListNode} l1 - The root node of list 1
- * @param {ListNode} l2 - The root node of list 2
+ * @param {ListNode} l1 - The root node of list 1.
+ * @param {ListNode} l2 - The root node of list 2.
+ * @returns {boolean} - true if has same data, false otherwise.
  */
 function hasSameData(l1, l2) {
   // end::description[]

book/interview-questions/longest-substring-without-repeating-characters.js

@@ -1,6 +1,16 @@
-// https://leetcode.com/problems/longest-substring-without-repeating-characters/submissions/
-
-function lengthOfLongestSubstring(s: string): number {
+// tag::description[]
+/**
+ * Find the length of the longest substring without duplicates.
+ * @example lenLongestSubstring('abccxyz'); // => 4 (cxyz)
+ * @param {string} s - The string.
+ * @returns {number} - The length of the longest unique substring.
+ */
+function lenLongestSubstring(s) {
+  // end::description[]
+  // tag::placeholder[]
+  // write your code here...
+  // end::placeholder[]
+  // tag::solution[]
   let max = 0;
   const set = new Set();
 
@@ -11,4 +21,9 @@ function lengthOfLongestSubstring(s: string): number {
   }
 
   return max;
-};
+  // end::solution[]
+  // tag::description[]
+}
+// end::description[]
+
+module.exports = { lenLongestSubstring };

book/interview-questions/longest-substring-without-repeating-characters.spec.js

@@ -1,5 +1,24 @@
-describe('', () => {
-  it('', () => {
+const { lenLongestSubstring } = require('./longest-substring-without-repeating-characters');
+// const {  } = require('../../src/index');
 
+[lenLongestSubstring].forEach((fn) => {
+  describe(`Set: ${fn.name}`, () => {
+    it('should work with null/empty', () => {
+      const actual = '';
+      const expected = 0;
+      expect(fn(actual)).toEqual(expected);
+    });
+
+    it('should work with small case', () => {
+      const actual = 'abc';
+      const expected = 3;
+      expect(fn(actual)).toEqual(expected);
+    });
+
+    it('should work with other case', () => {
+      const actual = 'abccdefg';
+      const expected = 5;
+      expect(fn(actual)).toEqual(expected);
+    });
   });
 });

book/interview-questions/max-subarray.js

@@ -7,6 +7,7 @@
  *  maxSubArray([-3,4,-1,2,1,-5]); // => 6
  *
  * @param {number[]} a - Array
+ * @returns {number} - max sum
  */
 function maxSubArray(a) {
 // end::description[]

book/interview-questions/merge-lists.js

@@ -9,6 +9,7 @@ const ListNode = require('../../src/data-structures/linked-lists/node');
  *
  * @param {ListNode} l1 - The root node of list 1
  * @param {ListNode} l2 - The root node of list 2
+ * @returns {ListNode} - The root of the merged list.
  */
 function mergeTwoLists(l1, l2) {
   // end::description[]