Hash Map vs Array vs Linked List for Lookup in DSA Python - Interactive Comparison Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to get the value for key 'apple' from the dictionary.

DSA Python

fruits = {'apple': 5, 'banana': 3, 'orange': 2}
value = fruits[1]'apple'
print(value)

Drag options to blanks, or click blank then click option'

A[

B.get(

C('

D{

Attempts:

3 left

💡 Hint

Common Mistakes

Using parentheses instead of square brackets for dictionary access.

Using dot notation which doesn't work for dictionary keys.

Use square brackets [] to access a value by key in a dictionary.

2fill in blank

medium

Complete the code to find the index of value 7 in the list.

DSA Python

numbers = [3, 7, 1, 9]
index = numbers.[1](7)
print(index)

Drag options to blanks, or click blank then click option'

Aindex

Bfind

Csearch

Dlocate

Attempts:

3 left

💡 Hint

Common Mistakes

Using methods like find which do not exist for lists.

Trying to use search or locate which are not list methods.

The index() method returns the first index of the given value in a list.

3fill in blank

hard

Fix the error in the linked list traversal to print all node values.

DSA Python

class Node:
    def __init__(self, val):
        self.val = val
        self.next = None

head = Node(1)
head.next = Node(2)
head.next.next = Node(3)

current = head
while current is not None:
    print(current[1]val)
    current = current.next

Drag options to blanks, or click blank then click option'

A:

B->

C.

D[

Attempts:

3 left

💡 Hint

Common Mistakes

Using arrow -> which is not valid in Python.

Using brackets or colon which are invalid for attribute access.

Use dot . to access attributes of an object in Python.

4fill in blank

hard

Fill both blanks to create a dictionary comprehension that maps words to their lengths for words longer than 3 letters.

DSA Python

words = ['cat', 'house', 'dog', 'elephant']
lengths = {word: [1] for word in words if len(word) [2] 3}
print(lengths)

Drag options to blanks, or click blank then click option'

Alen(word)

B<

C>

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using < instead of > for filtering.

Not using len(word) for the value.

Use len(word) to get the length and > to filter words longer than 3.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps uppercase words to their counts for counts greater than 0.

DSA Python

data = {'apple': 2, 'banana': 0, 'cherry': 5}
result = {{ [1]: [2] for k, v in data.items() if v [3] 0 }}
print(result)

Drag options to blanks, or click blank then click option'

Ak.upper()

Bv

C>

Dk.lower()

Attempts:

3 left

💡 Hint

Common Mistakes

Using k.lower() instead of k.upper().

Filtering with == or < instead of >.

Use k.upper() for uppercase keys, v for values, and filter with > 0.

Practice

1. Which data structure provides the fastest average lookup time for finding an element by key?

easy

A. Hash Map

B. Array

C. Linked List

D. Stack

Solution

1. Step 1: Understand lookup times for each data structure

   Hash Map uses a hash function to find keys quickly, usually in constant time O(1). Arrays require scanning or index access, O(1) if index known but O(n) if searching by value. Linked Lists require scanning nodes, O(n).

2. Step 2: Compare average lookup times

   Hash Map is fastest on average for key lookup due to direct hashing. Arrays and Linked Lists are slower for searching by value.

3. Final Answer:

   Hash Map -> Option A

4. Quick Check:

   Fastest lookup = Hash Map [OK]

Hint: Hash Map uses keys for O(1) lookup, arrays and lists do not [OK]

Common Mistakes:

• Confusing array index access with value lookup
• Thinking linked list lookup is fast
• Assuming stack is a lookup structure

2. Which of the following is the correct syntax to create an empty hash map (dictionary) in Python?

easy

A. hash_map = {}

B. hash_map = ()

C. hash_map = []

D. hash_map = set()

Solution

1. Step 1: Recall Python syntax for empty data structures

   Empty list uses [], empty tuple uses (), empty dictionary uses {}, and empty set uses set().

2. Step 2: Identify the hash map syntax

   Hash map in Python is a dictionary, created with {}.

3. Final Answer:

   hash_map = {} -> Option A

4. Quick Check:

   Empty dict = {} [OK]

Hint: Use curly braces {} for empty dictionary in Python [OK]

Common Mistakes:

• Using [] which creates a list, not a dictionary
• Using () which creates a tuple
• Using set() which creates a set, not a dictionary

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

arr = [10, 20, 30, 40]
lookup = {val: idx for idx, val in enumerate(arr)}
print(lookup.get(30))
print(lookup.get(50, -1))

medium

A. 30\n50

B. 3\n-1

C. 2\n-1

D. null\n-1

Solution

1. Step 1: Understand dictionary comprehension and get method

   The dictionary 'lookup' maps values to their indices: {10:0, 20:1, 30:2, 40:3}. lookup.get(30) returns the index for 30, which is 2. lookup.get(50, -1) looks for 50, not found, returns default -1.

2. Step 2: Determine printed output

   First print outputs 2. Second print outputs -1.

3. Final Answer:

   2 -1 -> Option C

4. Quick Check:

   lookup[30] = 2, missing key returns -1 [OK]

Hint: Dictionary get returns default if key missing [OK]

Common Mistakes:

• Confusing keys and values in dictionary
• Expecting get to raise error if key missing
• Mixing index and value in output

4. The following code tries to find if a value exists in a linked list. What is the bug?

class Node:
    def __init__(self, val):
        self.val = val
        self.next = None

def find(head, target):
    current = head
    while current.next is not None:
        if current.val == target:
            return True
        current = current.next
    return False

medium

A. The function should use recursion instead of a loop

B. The loop condition should be 'while current is not None' to check last node

C. The Node class is missing a previous pointer

D. The function should return false inside the loop

Solution

1. Step 1: Analyze loop condition in linked list traversal

   The loop uses 'while current.next is not None', so it stops before last node. This means the last node's value is never checked.

2. Step 2: Correct loop condition to include last node

   Changing to 'while current is not None' ensures all nodes, including last, are checked for target value.

3. Final Answer:

   The loop condition should be 'while current is not None' to check last node -> Option B

4. Quick Check:

   Loop must check all nodes including last [OK]

Hint: Check loop condition includes last node (current is not None) [OK]

Common Mistakes:

• Stopping loop early missing last node
• Adding unnecessary previous pointer
• Thinking recursion is required

5. You have a large dataset of unique user IDs and want to check membership quickly. Which data structure is best and why? A) Array because it stores elements sequentially B) Linked List because it allows easy insertion C) Hash Map because it provides average O(1) lookup D) Stack because it supports last-in-first-out access

hard

A. Array because it stores elements sequentially

B. Linked List because it allows easy insertion

C. Stack because it supports last-in-first-out access

D. Hash Map because it provides average O(1) lookup

Solution

1. Step 1: Understand lookup needs for large dataset

   Checking membership means quickly finding if an element exists. Arrays and linked lists require scanning, O(n) time. Stacks are not designed for membership lookup.

2. Step 2: Identify best data structure for fast membership

   Hash Map uses hashing to find elements in average O(1) time, ideal for large datasets and membership checks.

3. Final Answer:

   Hash Map because it provides average O(1) lookup -> Option D

4. Quick Check:

   Fast membership = Hash Map [OK]

Hint: Use hash map for fast membership checks in large data [OK]

Common Mistakes:

• Choosing array or list for fast lookup
• Confusing insertion speed with lookup speed
• Thinking stack supports membership lookup