String Reversal Approaches in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of slicing-based string reversal

What is the output of this Python code that reverses a string using slicing?

DSA Python

s = "hello"
reversed_s = s[::-1]
print(reversed_s)

A Error

B olleh

C hello

D h e l l o

💡 Hint

Think about how slicing with a step of -1 works on strings.

The slice s[::-1] creates a new string by taking characters from the end to the start, effectively reversing it.

❓ Predict Output

intermediate

Output of reversed() with join for string reversal

What is printed by this code that reverses a string using reversed() and join()?

DSA Python

s = "world"
reversed_s = ''.join(reversed(s))
print(reversed_s)

A world

B Error

C ['d', 'l', 'r', 'o', 'w']

D dlrow

💡 Hint

reversed() returns an iterator that goes backwards over the string.

reversed(s) returns characters from the end to the start. Joining them with an empty string creates the reversed string.

❓ Predict Output

advanced

Output of manual string reversal using a loop

What is the output of this code that reverses a string by building a new string in a loop?

DSA Python

s = "abcde"
result = ""
for char in s:
    result = char + result
print(result)

A Error

B abcde

C edcba

D aebcd

💡 Hint

Each new character is added before the existing result string.

By adding each character before the current result, the string is reversed step by step.

❓ Predict Output

advanced

Output of in-place reversal using list conversion

What is printed after reversing a string by converting it to a list and swapping characters in place?

DSA Python

s = "python"
chars = list(s)
left, right = 0, len(chars) - 1
while left < right:
    chars[left], chars[right] = chars[right], chars[left]
    left += 1
    right -= 1
reversed_s = ''.join(chars)
print(reversed_s)

A nohtyp

B python

C Error

D nothyp

💡 Hint

Swapping characters from ends towards the center reverses the list.

Swapping characters at left and right indices moves characters to reverse the list in place.

🧠 Conceptual

expert

Time complexity of different string reversal methods

Which string reversal approach has the best time complexity for reversing a string of length n?

A Converting string to list and swapping characters in place, O(n) time

B Using reversed() and join(), which is O(n) due to join optimization

C Using a loop that concatenates characters to a new string in each iteration, O(n²) time

D Using slicing s[::-1] which creates a new reversed string in O(n) time

💡 Hint

Consider how string concatenation inside loops affects time complexity.

Concatenating strings in a loop is costly because strings are immutable, causing O(n²) time. In-place list swapping and slicing are O(n). reversed() with join() is also O(n) because join is optimized.

Practice

1. Which of the following methods correctly reverses a string s in Python?

easy

A. Using reversed(s) without converting to string

B. Using slicing: s[::-1]

C. Using s.reverse() method

D. Using s.append() method

Solution

1. Step 1: Understand string slicing

   The slicing syntax s[::-1] creates a new string with characters in reverse order.

2. Step 2: Check other methods

   s.reverse() and s.append() are list methods, not valid for strings. reversed(s) returns an iterator, not a string, so it needs conversion.

3. Final Answer:

   Using slicing: s[::-1] -> Option B

4. Quick Check:

   String slicing with negative step reverses string [OK]

Hint: Use s[::-1] to reverse strings quickly [OK]

Common Mistakes:

• Trying to use list methods on strings
• Forgetting to convert reversed() iterator to string
• Using append() on strings

2. Which of the following code snippets correctly reverses the string text = 'hello' and prints it?

easy

A. print(text.append('o'))

B. print(text.reverse())

C. print(text[::-2])

D. print(''.join(reversed(text)))

Solution

1. Step 1: Analyze each option

   text.reverse() is invalid for strings. text[::-2] reverses with step 2, skipping characters. text.append() is invalid for strings.

2. Step 2: Confirm correct syntax

   reversed(text) returns an iterator of characters reversed. ''.join() converts it back to string. This prints 'olleh'.

3. Final Answer:

   print(''.join(reversed(text))) -> Option D

4. Quick Check:

   reversed() + join() prints reversed string [OK]

Hint: Use ''.join(reversed(s)) to reverse strings [OK]

Common Mistakes:

• Using reverse() method on strings
• Using slicing with wrong step
• Trying to append to strings

3. What is the output of the following code?

def reverse_string(s):
    result = ''
    for char in s:
        result = char + result
    return result

print(reverse_string('abc'))

medium

A. 'cba'

B. 'abc'

C. 'bac'

D. Error

Solution

1. Step 1: Trace the loop for s='abc'

   Initially, result = ''. Iteration 1: char='a', result='a' + '' = 'a'. Iteration 2: char='b', result='b' + 'a' = 'ba'. Iteration 3: char='c', result='c' + 'ba' = 'cba'.

2. Step 2: Return the reversed string

   The function returns 'cba', which is the reverse of 'abc'.

3. Final Answer:

   'cba' -> Option A

4. Quick Check:

   Prepending chars reverses string [OK]

Hint: Add each char before result to reverse string [OK]

Common Mistakes:

• Appending char instead of prepending
• Returning original string
• Confusing order of concatenation

4. The following code is intended to reverse a string but has an error. What is the error?

def reverse_str(s):
    s.reverse()
    return s

print(reverse_str('hello'))

medium

A. Missing return statement

B. Function returns None

C. Strings do not have reverse() method

D. Syntax error in function definition

Solution

1. Step 1: Check method usage on string

   The reverse() method exists for lists, not strings. Calling it on a string causes AttributeError.

2. Step 2: Understand the error cause

   Since strings are immutable, they do not support reverse(). The code will raise an error at runtime.

3. Final Answer:

   Strings do not have reverse() method -> Option C

4. Quick Check:

   reverse() is list method, not string method [OK]

Hint: Use slicing or reversed() for strings, not reverse() [OK]

Common Mistakes:

• Trying to use reverse() on strings
• Expecting reverse() to modify string in place
• Ignoring immutability of strings

5. You want to reverse only the vowels in a string s = 'hello world' while keeping other characters in place. Which approach below correctly achieves this?

hard

A. Extract vowels, reverse them, then replace vowels in original positions

B. Reverse entire string using s[::-1]

C. Use reversed(s) and join all characters

D. Replace vowels with '*' and then reverse string

Solution

1. Step 1: Understand the problem

   We want to reverse only vowels, keeping consonants and spaces fixed in their places.

2. Step 2: Analyze each option

   Reverse entire string using s[::-1] reverses whole string, changing all characters. Use reversed(s) and join all characters reverses all characters, not just vowels. Replace vowels with '*' and then reverse string replaces vowels with '*', losing original vowels. Extract vowels, reverse them, then replace vowels in original positions extracts vowels, reverses them, then puts them back in original vowel positions, preserving other characters.

3. Final Answer:

   Extract vowels, reverse them, then replace vowels in original positions -> Option A

4. Quick Check:

   Reverse vowels only by extracting and replacing [OK]

Hint: Extract vowels, reverse, then put back in place [OK]

Common Mistakes:

• Reversing entire string instead of vowels only
• Replacing vowels with placeholders without restoring
• Ignoring character positions