Hash map vs hash set in Data Structures Theory - Practice Questions

Challenge - 5 Problems

🧠 Conceptual

intermediate

Difference in data storage between hash map and hash set

Which statement correctly describes the difference between a hash map and a hash set?

A A hash map stores key-value pairs, while a hash set stores only unique keys without associated values.

B Both hash map and hash set store key-value pairs but differ in sorting order.

C A hash map stores only unique keys, while a hash set stores key-value pairs.

D A hash map stores values only, while a hash set stores keys only.

💡 Hint

Think about whether each structure associates values with keys or just stores keys.

A hash map stores pairs of keys and their associated values, allowing lookup by key. A hash set stores only unique keys without any associated values.

📋 Factual

intermediate

Typical use case for a hash set

Which of the following is a common use case for a hash set?

A Checking if an item has already been seen to avoid duplicates.

B Storing ordered pairs of data.

C Mapping user IDs to user profiles.

D Counting the frequency of words in a text.

💡 Hint

Consider when you only need to know if something exists or not.

Hash sets are used to store unique items and quickly check if an item is already present, which helps avoid duplicates.

🔍 Analysis

advanced

Performance difference in lookup operations

Given a hash map and a hash set both implemented with good hashing functions, which statement about their lookup performance is true?

A Lookup in a hash map is generally slower than in a hash set because it retrieves values, not just keys.

B Lookup time complexity is generally the same (average O(1)) for both hash map and hash set.

C Hash sets have slower lookup because they store only keys without values.

D Hash maps have O(n) lookup time while hash sets have O(1).

💡 Hint

Think about how hashing works for both structures.

Both hash maps and hash sets use hashing to achieve average constant time (O(1)) lookup for keys. The presence of values in hash maps does not affect the lookup time complexity.

❓ Comparison

advanced

Memory usage comparison between hash map and hash set

Which statement best describes the memory usage difference between a hash map and a hash set storing the same number of keys?

A A hash set uses more memory because it stores keys twice for faster lookup.

B A hash map uses less memory because it compresses values.

C Both use the same amount of memory since they store the same keys.

D A hash map uses more memory because it stores both keys and values, while a hash set stores only keys.

💡 Hint

Consider what extra data each structure holds besides keys.

Hash maps store key-value pairs, so they require memory for both keys and values. Hash sets store only keys, so they use less memory for the same number of keys.

❓ Reasoning

expert

Choosing between hash map and hash set for a problem

You need to track unique user IDs that have logged into a system and also store the last login timestamp for each user. Which data structure is the best choice?

A Use a hash set to store user IDs and a separate list to store timestamps.

B Use a hash set only, storing timestamps as keys.

C Use a hash map with user IDs as keys and timestamps as values.

D Use a list of tuples containing user IDs and timestamps.

💡 Hint

Think about how to associate each user ID with its timestamp efficiently.

A hash map allows storing each user ID as a key with its last login timestamp as the associated value, enabling quick lookup and update. A hash set cannot store associated values.

Practice

1. Which data structure is best suited for storing unique items without any associated values?

easy

A. Hash set

B. Linked list

C. Array

D. Hash map

Solution

1. Step 1: Understand the purpose of a hash set

   A hash set stores only unique values and does not associate keys with values.

2. Step 2: Compare with hash map

   A hash map stores key-value pairs, so it is not just for unique values but for mapping keys to values.

3. Final Answer:

   Hash set -> Option A

4. Quick Check:

   Unique values only = Hash set [OK]

Hint: Unique items without keys? Think hash set [OK]

Common Mistakes:

• Confusing hash map with hash set
• Thinking hash map stores unique values only
• Assuming arrays enforce uniqueness

2. Which of the following is the correct way to declare a hash map in Python?

easy

A. my_map = ["key1", "value1", "key2", "value2"]

B. my_map = ("key1", "value1", "key2", "value2")

C. my_map = {"key1": "value1", "key2": "value2"}

D. my_map = set("key1", "key2")

Solution

1. Step 1: Recognize Python dictionary syntax

   Python hash maps are dictionaries declared with curly braces and key-value pairs separated by colons.

2. Step 2: Check each option

   my_map = {"key1": "value1", "key2": "value2"} uses curly braces and key-value pairs correctly. Options A and C use lists or tuples, not dictionaries. my_map = set("key1", "key2") creates a set, not a map.

3. Final Answer:

   my_map = {"key1": "value1", "key2": "value2"} -> Option C

4. Quick Check:

   Curly braces with key:value = dict [OK]

Hint: Curly braces with colons mean hash map in Python [OK]

Common Mistakes:

• Using lists or tuples instead of dictionaries
• Confusing sets with maps
• Missing colons between keys and values

3. What will be the output of the following Python code?

my_set = {1, 2, 3}
my_set.add(2)
print(len(my_set))

medium

A. 4

B. 3

C. 2

D. Error

Solution

1. Step 1: Understand set behavior on adding duplicates

   Sets store unique values. Adding 2 again does not increase size.

2. Step 2: Calculate length after adding duplicate

   Initial set has 3 elements. Adding 2 again keeps it at 3.

3. Final Answer:

   3 -> Option B

4. Quick Check:

   Set ignores duplicates = length unchanged [OK]

Hint: Adding duplicate to set doesn't increase size [OK]

Common Mistakes:

• Assuming duplicates increase set size
• Confusing set with list behavior
• Expecting an error on duplicate add

4. Identify the error in this JavaScript code snippet that tries to use a hash set:

const mySet = new Set();
mySet.add('apple');
mySet.add('banana');
mySet.add('apple');
console.log(mySet.length);

medium

A. Using 'length' property instead of 'size' for Set

B. Adding duplicate values to Set causes error

C. Set cannot store strings

D. Missing parentheses in Set constructor

Solution

1. Step 1: Check Set properties in JavaScript

   JavaScript Sets use 'size' property to get number of elements, not 'length'.

2. Step 2: Analyze the code

   Code uses 'mySet.length' which is undefined, causing incorrect output.

3. Final Answer:

   Using 'length' property instead of 'size' for Set -> Option A

4. Quick Check:

   Set size = size property, not length [OK]

Hint: Use .size, not .length, to get Set size in JavaScript [OK]

Common Mistakes:

• Using length instead of size for Sets
• Thinking duplicates cause errors
• Believing Sets can't store strings

5. You have a list of student names with possible duplicates. You want to count how many unique students there are and also store their grades. Which data structures should you use together?

hard

A. Use two hash sets: one for names and one for grades

B. Use only a hash map with names as keys and grades as values

C. Use a hash set for grades and a hash map for names

D. Use a hash set for unique names and a hash map for name-grade pairs

Solution

1. Step 1: Identify need for uniqueness and mapping

   Counting unique students requires a hash set to store unique names. Storing grades requires a hash map to map each name to a grade.

2. Step 2: Evaluate options

   Use a hash set for unique names and a hash map for name-grade pairs uses both structures correctly. Use only a hash map with names as keys and grades as values does not separate uniqueness check explicitly. Options C and D misuse sets and maps.

3. Final Answer:

   Use a hash set for unique names and a hash map for name-grade pairs -> Option D

4. Quick Check:

   Uniqueness + mapping = set + map [OK]

Hint: Unique names = set; name to grade = map [OK]

Common Mistakes:

• Using only one data structure for both tasks
• Confusing sets for storing grades
• Not separating uniqueness from mapping