Palindrome Detection in DSA Python

Mental Model

Check if a sequence reads the same forwards and backwards by comparing characters from both ends moving towards the center.

Analogy: Like checking if a word on a mirror looks the same on both sides, such as 'madam' or 'racecar'.

start -> [m] a d a m ← end
indexes: 0 -> 1 -> 2 -> 3 -> 4

Dry Run Walkthrough

Input: string: 'radar'

Goal: Determine if 'radar' reads the same forwards and backwards

Step 1: Compare first and last characters 'r' and 'r'

r a d a r
↑           ↑
left=0     right=4

Why: Check if characters at both ends match to confirm palindrome property

Step 2: Move inward and compare 'a' and 'a'

r a d a r
  ↑     ↑
left=1 right=3

Why: Continue checking matching pairs moving towards the center

Step 3: Move inward to middle character 'd'

r a d a r
    ↑
left=2 right=2

Why: Middle character doesn't need a pair, so palindrome confirmed

Result:

r a d a r
Palindrome: True

Annotated Code

DSA Python

class PalindromeDetector:
    def is_palindrome(self, s: str) -> bool:
        left, right = 0, len(s) - 1
        while left < right:
            if s[left] != s[right]:
                return False
            left += 1
            right -= 1
        return True

if __name__ == '__main__':
    detector = PalindromeDetector()
    test_string = 'radar'
    result = detector.is_palindrome(test_string)
    print(f"{test_string} -> Palindrome: {result}")

while left < right:

loop to compare characters from both ends moving inward

if s[left] != s[right]:

check if characters at current pointers differ, which breaks palindrome

left += 1 right -= 1

advance pointers inward to next characters for comparison

return True

all pairs matched, confirm string is palindrome

Output

radar -> Palindrome: True

Complexity Analysis

Time: O(n) because we compare pairs of characters from both ends once, where n is string length

Space: O(1) because only two pointers are used regardless of input size

vs Alternative: Compared to reversing the string and comparing, this method uses less space and stops early on mismatch

Edge Cases

empty string

returns True since empty string reads the same forwards and backwards

DSA Python

while left < right:

single character string

returns True because a single character is always a palindrome

DSA Python

while left < right:

string with different characters

returns False as soon as a mismatch is found

DSA Python

if s[left] != s[right]:

When to Use This Pattern

When you need to check if a sequence reads the same forwards and backwards, use two pointers moving inward to compare pairs efficiently.

Common Mistakes

Mistake: Comparing characters only from start to middle without checking the corresponding end character
Fix: Use two pointers, one at start and one at end, and compare both simultaneously

Mistake: Reversing the string and comparing without considering extra space
Fix: Use two-pointer approach to save space and allow early exit on mismatch

Summary

Checks if a string reads the same forwards and backwards by comparing characters from both ends.

Use when you want to verify if a word or sequence is a palindrome efficiently.

The key insight is to use two pointers moving inward to compare pairs, stopping early if mismatch occurs.