Implement is_palindrome(s)
Write a recursive Python function that returns True if a string reads the same forwards and backwards, and False otherwise. The trick is to compare the outer two characters and recurse on everything in between.
The problem
A string is a palindrome if its first and last characters match and the substring in between is also a palindrome. Strings of length 0 or 1 are trivially palindromes — those are your base cases.
"r" == "r" → True, recurse on "aceca"
"a" == "a" → True, recurse on "cec"
"c" == "c" → True, recurse on "e"
len("e") <= 1 → True ← base case
Your function must pass these tests:
- is_palindrome("") → True
- is_palindrome("a") → True
- is_palindrome("racecar") → True
- is_palindrome("hello") → False
- is_palindrome("level") → True
Write your solution below, then click Run Tests:
Hints
Hint 1 — Negative indices in Python
Python allows negative indices: s[-1] is the last character, equivalent to s[len(s) - 1]. The slice s[1:-1] gives you everything except the first and last character.
Hint 2 — Short-circuiting with and
If the outer characters do not match, the string cannot be a palindrome — you can return False immediately without recursing. Using and handles this automatically: s[0] == s[-1] and is_palindrome(s[1:-1]).
Hint 3 — Full solution
def is_palindrome(s):
if len(s) <= 1:
return True
return s[0] == s[-1] and is_palindrome(s[1:-1])The and operator short-circuits: if s[0] != s[-1], it returns False without making the recursive call.