Why Strings Are a Data Structure Not Just Text in DSA Python - Test Your Knowledge

Practice - 5 Tasks

1fill in blank

easy

Complete the code to get the first character of the string.

DSA Python

word = "hello"
first_char = word[1]

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

A<0>

B(0)

C[0]

D{0}

Attempts:

3 left

💡 Hint

Common Mistakes

Using parentheses instead of square brackets

Starting index at 1

The first character of a string is accessed using square brackets with index 0.

2fill in blank

medium

Complete the code to find the length of the string.

DSA Python

word = "hello"
length = [1](word)

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

Alength

Bsize

Ccount

Dlen

Attempts:

3 left

💡 Hint

Common Mistakes

Using non-existent functions like size() or length()

The len() function returns the number of characters in a string.

3fill in blank

hard

Fix the error in the code to get the last character of the string.

DSA Python

word = "hello"
last_char = word[1]

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

A[1]

B[-1]

C[5]

D[-5]

Attempts:

3 left

💡 Hint

Common Mistakes

Using positive indexes beyond string length

Using index 1 which is the second character

Using index -1 accesses the last character of the string.

4fill in blank

hard

Fill both blanks to create a dictionary with words as keys and their lengths as values.

DSA Python

words = ["apple", "banana", "cherry"]
lengths = {word[1] for word in words if len(word) [2] 5}

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

A: len(word)

B>

C==

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using '=' instead of ':' in dictionary comprehension

Using wrong comparison operators

The dictionary comprehension uses word: len(word) to map words to their lengths, and filters words with length greater than 5.

5fill in blank

hard

Fill all three blanks to create a dictionary with uppercase words as keys and their lengths as values, filtering words longer than 5.

DSA Python

words = ["apple", "banana", "cherry"]
result = { [1]: [2] for word in words if len(word) [3] 5 }

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

Aword.upper()

Blen(word)

C>

Dword.lower()

Attempts:

3 left

💡 Hint

Common Mistakes

Using lowercase instead of uppercase keys

Using wrong comparison operators

The dictionary comprehension uses word.upper() as keys, len(word) as values, and filters words with length greater than 5.

Practice

1. Why are strings considered a data structure and not just plain text in programming?

easy

A. Because strings are only used for printing messages

B. Because strings only store numbers

C. Because strings store a sequence of characters that can be accessed and manipulated individually

D. Because strings cannot be changed once created

Solution

1. Step 1: Understand what a string stores

   A string holds many characters in order, like a list of letters.

2. Step 2: Recognize string operations

   You can access, change, or count characters, showing it acts like a data structure.

3. Final Answer:

   Because strings store a sequence of characters that can be accessed and manipulated individually -> Option C

4. Quick Check:

   Strings = sequence of characters [OK]

Hint: Strings hold characters in order, like a list [OK]

Common Mistakes:

• Thinking strings only hold text without structure
• Confusing immutability with data structure nature
• Believing strings only store numbers

2. Which of the following is the correct way to access the third character of a string s = 'hello' in Python?

easy

A. s[3]

B. s(3)

C. s{3}

D. s[2]

Solution

1. Step 1: Recall Python string indexing

   Python strings use zero-based indexing, so first character is s[0].

2. Step 2: Identify third character index

   The third character is at index 2, so s[2] accesses it.

3. Final Answer:

   s[2] -> Option D

4. Quick Check:

   Third char = s[2] [OK]

Hint: Remember Python indexes start at 0 [OK]

Common Mistakes:

• Using parentheses instead of brackets
• Using 1-based indexing
• Using curly braces for indexing

3. What will be the output of this code?

text = 'data'
print(text[1:3])

medium

A. ta

B. at

C. da

D. ad

Solution

1. Step 1: Understand string slicing syntax

   text[1:3] means characters from index 1 up to but not including 3.

2. Step 2: Identify characters at indexes 1 and 2

   text = 'd'(0), 'a'(1), 't'(2), 'a'(3). So indexes 1 and 2 are 'a' and 't'.

3. Final Answer:

   at -> Option B

4. Quick Check:

   Slicing text[1:3] = at [OK]

Hint: Slice excludes end index, so stop before 3 [OK]

Common Mistakes:

• Including the character at index 3
• Confusing start and end indexes
• Thinking slicing returns a list

4. Identify the error in this code snippet:

word = 'hello'
word[0] = 'H'
print(word)

medium

A. Strings are immutable; you cannot assign to an index

B. Syntax error due to missing quotes

C. print statement is incorrect

D. No error; output will be 'Hello'

Solution

1. Step 1: Understand string immutability

   Strings cannot be changed by assigning to an index; they are immutable.

2. Step 2: Check the assignment line

   word[0] = 'H' tries to change first character, which causes an error.

3. Final Answer:

   Strings are immutable; you cannot assign to an index -> Option A

4. Quick Check:

   Immutable strings disallow index assignment [OK]

Hint: Strings cannot be changed by index assignment [OK]

Common Mistakes:

• Thinking strings can be changed like lists
• Ignoring error message
• Assuming print causes error

5. Given the string s = 'abracadabra', which code snippet correctly counts how many times the letter 'a' appears?

hard

A. All of the above

B. count = s.count('a')
print(count)

C. count = 0
for i in range(len(s)):
if s[i] == 'a':
count += 1
print(count)

D. count = 0
for ch in s:
if ch == 'a':
count += 1
print(count)

Solution

1. Step 1: Analyze each code snippet

   count = 0 for ch in s: if ch == 'a': count += 1 print(count) uses a loop and manual count; count = s.count('a') print(count) uses built-in count method; count = 0 for i in range(len(s)): if s[i] == 'a': count += 1 print(count) uses index loop and count.

2. Step 2: Verify correctness of all methods

   All three correctly count 'a' occurrences in the string.

3. Final Answer:

   All of the above -> Option A

4. Quick Check:

   All methods count 'a' correctly [OK]

Hint: Built-in count or loops both work to count chars [OK]

Common Mistakes:

• Using wrong loop indexes
• Forgetting to increment count
• Assuming only built-in method works