Palindrome Detection in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if a string is the same forwards and backwards.

DSA Python

def is_palindrome(s):
    return s == s[1]

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

A[::-1]

B[1:]

C[:1]

D[::-2]

Attempts:

3 left

💡 Hint

Common Mistakes

Using [1:] or [:1] which do not reverse the string.

Using [::-1] reverses the string, so comparing s to s[::-1] checks if it reads the same backwards.

2fill in blank

medium

Complete the code to ignore case when checking palindrome.

DSA Python

def is_palindrome(s):
    s = s.lower()
    return s == s[1]

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

A[1:]

B[::-1]

C[:1]

D[::2]

Attempts:

3 left

💡 Hint

Common Mistakes

Not reversing the string correctly after lowering case.

After converting to lowercase, reversing with [::-1] allows case-insensitive palindrome check.

3fill in blank

hard

Fix the error in the code to correctly check palindrome ignoring spaces.

DSA Python

def is_palindrome(s):
    s = s.replace(' ', '')
    return s == s[1]

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

A[::-1]

B[1:]

C[:1]

D[::2]

Attempts:

3 left

💡 Hint

Common Mistakes

Using incorrect slicing that does not reverse the string.

Removing spaces and then reversing with [::-1] checks palindrome ignoring spaces.

4fill in blank

hard

Fill both blanks to create a palindrome check ignoring case and non-alphanumeric characters.

DSA Python

import re

def is_palindrome(s):
    s = re.sub([1], '', s).lower()
    return s == s[2]

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

A'[^a-zA-Z0-9]'

B' '

C[::-1]

D[1:]

Attempts:

3 left

💡 Hint

Common Mistakes

Removing only spaces instead of all non-alphanumeric characters.

Using wrong slice to reverse.

Using regex '[^a-zA-Z0-9]' removes all non-alphanumeric characters. Then reversing with [::-1] checks palindrome ignoring case and special characters.

5fill in blank

hard

Fill all three blanks to create a palindrome check ignoring case, spaces, and punctuation.

DSA Python

import string

def is_palindrome(s):
    s = ''.join(c for c in s.lower() if c not in [1])
    return s == s[2] and len(s) > [3]

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

Astring.punctuation + ' '

B[::-1]

C0

Dstring.whitespace

Attempts:

3 left

💡 Hint

Common Mistakes

Not removing all punctuation and spaces.

Using wrong slice to reverse.

Not checking string length.

Removing punctuation and spaces using string.punctuation + ' ', reversing with [::-1], and checking length greater than 0 ensures a valid palindrome check.

Practice

1. What does it mean if a string is a palindrome?

easy

A. It contains only letters

B. It reads the same forwards and backwards

C. It has an even number of characters

D. It starts and ends with the same letter

Solution

1. Step 1: Understand palindrome definition

   A palindrome is a word or phrase that reads the same forwards and backwards.

2. Step 2: Compare options with definition

   Only It reads the same forwards and backwards matches this definition exactly.

3. Final Answer:

   It reads the same forwards and backwards -> Option B

4. Quick Check:

   Palindrome = Same forwards and backwards [OK]

Hint: Check if string equals its reverse [OK]

Common Mistakes:

• Thinking palindrome means only letters
• Assuming even length only
• Confusing start/end letters with palindrome

2. Which of the following Python code snippets correctly checks if a string s is a palindrome?

easy

A. s == reversed(s)

B. s == s[1:]

C. s == s.reverse()

D. s == s[::-1]

Solution

1. Step 1: Understand string reversal in Python

   The slice s[::-1] returns the reversed string.

2. Step 2: Check each option

   s == s[::-1] compares s to its reversed slice correctly. s == reversed(s) compares to an iterator, not a string. s == s.reverse() tries to use reverse() which is a list method and returns None. s == s[1:] compares to a substring missing the first character.

3. Final Answer:

   s == s[::-1] -> Option D

4. Quick Check:

   Use slicing to reverse string [OK]

Hint: Use s == s[::-1] to check palindrome [OK]

Common Mistakes:

• Using reversed() without join
• Calling reverse() on string
• Comparing to wrong substring

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

def is_palindrome(s):
    return s == s[::-1]

print(is_palindrome('level'))
print(is_palindrome('world'))

medium

A. True\nFalse

B. False\nTrue

C. True\nTrue

D. False\nFalse

Solution

1. Step 1: Check 'level' palindrome status

   'level' reversed is 'level', so is_palindrome('level') returns True.

2. Step 2: Check 'world' palindrome status

   'world' reversed is 'dlrow', which is different, so is_palindrome('world') returns False.

3. Final Answer:

   True\nFalse -> Option A

4. Quick Check:

   'level' palindrome True, 'world' not [OK]

Hint: Check string equals reversed string for each input [OK]

Common Mistakes:

• Assuming all strings are palindrome
• Mixing output order
• Confusing print output format

4. Find the error in this palindrome check function:

def check_palindrome(s):
    for i in range(len(s)):
        if s[i] != s[len(s)-i]:
            return False
    return True

medium

A. Function returns True too early

B. Loop should run only half the string length

C. Index out of range error in s[len(s)-i]

D. No error, function is correct

Solution

1. Step 1: Analyze index usage in loop

   Index len(s)-i is out of range because string indices go from 0 to len(s)-1.

2. Step 2: Identify correct index for reverse comparison

   It should be len(s)-i-1 to avoid index error.

3. Final Answer:

   Index out of range error in s[len(s)-i] -> Option C

4. Quick Check:

   Check index bounds carefully [OK]

Hint: Remember last index is len(s)-1 [OK]

Common Mistakes:

• Off-by-one error in indexing
• Looping full length instead of half
• Ignoring index errors

5. You want to check if a phrase is a palindrome ignoring spaces and capitalization. Which code snippet correctly implements this?

hard

A. s_clean = ''.join(c.lower() for c in s if c.isalpha())\nreturn s_clean == s_clean[::-1]

B. s_clean = s.replace(' ', '')\nreturn s_clean == s_clean[::-1]

C. s_clean = s.lower()\nreturn s_clean == s_clean[::-1]

D. s_clean = ''.join(c for c in s if c.isdigit())\nreturn s_clean == s_clean[::-1]

Solution

1. Step 1: Understand requirements

   We must ignore spaces and capitalization, and consider only letters.

2. Step 2: Analyze each option

   s_clean = ''.join(c.lower() for c in s if c.isalpha())\nreturn s_clean == s_clean[::-1] removes non-letters and lowercases all characters, then checks palindrome. s_clean = s.replace(' ', '')\nreturn s_clean == s_clean[::-1] only removes spaces but keeps punctuation and case. s_clean = s.lower()\nreturn s_clean == s_clean[::-1] keeps punctuation and spaces. s_clean = ''.join(c for c in s if c.isdigit())\nreturn s_clean == s_clean[::-1] keeps digits only, ignoring letters.

3. Final Answer:

   s_clean = ''.join(c.lower() for c in s if c.isalpha())\nreturn s_clean == s_clean[::-1] -> Option A

4. Quick Check:

   Filter letters and lowercase before palindrome check [OK]

Hint: Filter letters and lowercase before checking [OK]

Common Mistakes:

• Only removing spaces
• Ignoring punctuation
• Not lowercasing letters