Palindrome Detection in DSA Python - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a palindrome?

A palindrome is a word, number, phrase, or sequence that reads the same backward as forward, ignoring spaces, punctuation, and capitalization.

beginner

How can you check if a string is a palindrome using two pointers?

Use two pointers: one at the start and one at the end of the string. Move both pointers towards the center, comparing characters. If all pairs match, it's a palindrome.

beginner

Why do we ignore spaces and punctuation when detecting palindromes?

Ignoring spaces and punctuation helps focus on the actual letters or digits, making the palindrome check meaningful for phrases and sentences, not just single words.

intermediate

What is the time complexity of checking a palindrome in a string of length n?

The time complexity is O(n) because each character is checked at most once when comparing pairs from start and end.

beginner

Show a simple Python code snippet to check if a string is a palindrome ignoring case and non-alphanumeric characters.

import re

def is_palindrome(s: str) -> bool:
    s = re.sub(r'[^a-zA-Z0-9]', '', s).lower()
    return s == s[::-1]

Which of the following strings is a palindrome?

A racecar

B hello

C world

D python

The word 'racecar' reads the same backward and forward.

What is the main idea behind the two-pointer technique for palindrome detection?

A Sort the string and check if it matches the original

B Compare characters from the start and end moving towards the center

C Reverse the string and compare with the original

D Count the frequency of each character

Two pointers start at opposite ends and move inward comparing characters.

Why do we convert the string to lowercase before checking palindrome?

A To remove punctuation

B To speed up the process

C To ignore case differences

D To remove spaces

Lowercasing ensures 'A' and 'a' are treated the same.

What is the time complexity of palindrome detection using the two-pointer method?

A O(1)

B O(n^2)

C O(log n)

D O(n)

Each character is checked once, so time is linear.

Which Python method helps to reverse a string easily?

A s[::-1]

B s.reverse()

C reversed(s)

D s.reverseString()

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

Explain how to detect if a string is a palindrome using the two-pointer technique.

Think about checking characters from both ends moving inward.

Describe why preprocessing (like removing spaces and punctuation) is important in palindrome detection for sentences.

Consider how phrases differ from single words.

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