Check a palindrome recursively by indices

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Problem

Implement is_palindrome(text) recursively without loops, slicing, or reversing. The comparison is case-sensitive and includes every character.

Reveal answer or reference solution

Use two indices. Zero or one remaining characters is a palindrome. If the endpoint characters differ, return false; otherwise recurse inward. The algorithm takes $O(n)$ time and $O(n)$ call-stack space.

def is_palindrome(text: str) -> bool:
    def check(left: int, right: int) -> bool:
        if left >= right:
            return True
        if text[left] != text[right]:
            return False
        return check(left + 1, right - 1)
 
    return check(0, len(text) - 1)

Public tests

is_palindrome("") → True
is_palindrome("x") → True
is_palindrome("racecar") → True
is_palindrome("abca") → False
is_palindrome("Aa") → False

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Check a palindrome recursively by indices

Problem

Public tests

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