Palindrome String | Recursive & Iterative Python Solutions

Learn two easy ways to test a Palindrome String: a clean recursive method and a space saving two-pointer loop. Includes code with comments, step-by-step explanation, dry run, and Big-O analysis.

The [Problem Link] is given here for your quick reference.

What does the problem ask?

Given a string s, tell whether it reads the same forwards and backwards.
Return True if it is a palindrome, or False otherwise.

Examples

Intuition & Approach

Key idea for both methods

Compare matching characters from the left and right ends and move inwards:

• If every pair matches → palindrome.
• If any pair differs → not a palindrome.

We show two ways to do this:

1. Recursive – a clean self-calling check.
2. Iterative (two pointers) – a simple while loop.

Code 1 – Recursive

class Solution:
    # helper that checks s[left .. right]
    def solve(self, s, left, right):
        if left >= right:        # crossed over → all pairs matched
            return True
        if s[left] != s[right]:  # mismatch
            return False
        # move both pointers inward
        return self.solve(s, left + 1, right - 1)

    def isPalindrome(self, s: str) -> bool:
        # check the whole string
        return self.solve(s, 0, len(s) - 1)

Code 2 – Iterative Two Pointers

class Solution:
    def isPalindrome(self, s: str) -> bool:
        n = len(s)
        left = 0
        right = n - 1
        while left < right:
            if s[left] != s[right]:   # mismatch
                return False
            left += 1                 # move inward
            right -= 1
        return True                   # all pairs matched

Code Walkthrough

Recursive method

• Base cases
  • left >= right → checked all pairs → return True.
  • Characters differ → return False.
• Recursive step
  Move left rightward and right leftward and call again.

Iterative method

• Set two pointers: left at start, right at end.
• While they have not crossed:
  • If characters differ → not a palindrome.
  • Else move both pointers inward.
• Loop ends without a mismatch → palindrome.

Dry Run (iterative method on "abba")

Complexity

n is the length of the string.

Conclusion

A palindrome check needs only a linear scan from both ends.

• The recursive version is short and clear but uses extra stack space.
• The iterative two-pointer version saves space and is often preferred in practice.

Choose the one that best fits your style or memory limits.

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