Date: 2026-06-06
Time: 19:38
This file implements LeetCode 680 — Valid Palindrome II. Given a string, it determines whether the string can become a palindrome by deleting at most one character. It exports a single function validPalindrome used by the test suite.
validPalindrome(s: str) -> boolThe top-level solver. Uses a two-pointer approach that walks inward from both ends of the string. When a mismatch is found, it forks into two possibilities — skip the left character or skip the right character — and checks if either produces a palindrome. Only one mismatch is tolerated, which enforces the "at most one deletion" constraint.
is_palindrome(left: int, right: int) -> boolA closure over s that checks whether the substring s[left:right+1] is a palindrome using a simple inward sweep. It's called at most twice (on mismatch), and only over the remaining suffix of the string — not the full string — so it doesn't redo work already validated by the outer loop.
Two-pointer palindrome check — the canonical O(n) approach. The outer loop narrows the window; is_palindrome handles the residual check after a skip.
Greedy single-fork — on the first mismatch, both skip options are tried via or. If either works, the string qualifies. There's no backtracking or recursion beyond this single fork, which keeps it O(n) overall.
Closure instead of parameter passing — is_palindrome captures s from the enclosing scope rather than taking it as an argument, reducing boilerplate.
Imports: None. Pure Python, no standard library usage.
Imported by: valid-palindrome-ii/test_solution.py directly. The massive "Imported By" list in the prompt is an artifact of the repo's test infrastructure — those other test files don't actually import this solution; they share a common test runner pattern.
1. Initialize left = 0, right = len(s) - 1.
2. Walk inward: if s[left] == s[right], advance both pointers.
3. On first mismatch: return ispalindrome(left+1, right) or ispalindrome(left, right-1). This is the "use your one deletion" step.
4. If the outer loop completes without mismatch, the string is already a palindrome — return True.
The key insight: characters validated by the outer loop before the mismatch are guaranteed to be symmetric, so is_palindrome only needs to check the remaining inner substring.
is_palindrome). No auxiliary data structures.s contains only lowercase English letters (per LeetCode constraints). No validation is performed.None. The function assumes valid input — a non-empty string of lowercase letters. It won't crash on empty strings (the while left < right guard handles it), but there's no explicit validation or exception raising.
valid-palindrome/solution.py — The simpler predecessor problem (no deletions allowed); compare to see how the deletion fork extends the basic patternvalid-palindrome-ii/solution.py:is_palindrome — Verify that it's never called on the full string, only on the reduced window after a mismatchvalid-palindrome-ii/test_solution.py — Check which edge cases are covered (empty string, single char, already-palindrome, deletion at start vs end vs middle)greedy-vs-dp-palindrome — Why a greedy single-fork works for k=1 deletions but wouldn't generalize to k>1 without DPat-most-two-is-palindrome-calls — is_palindrome is called at most twice per invocation of validPalindrome, and only when the outer loop encounters a mismatchlinear-time-guarantee — The total work across the outer loop and both is_palindrome calls is bounded by 2n character comparisonsno-allocation — The solution uses zero auxiliary data structures; space complexity is O(1) excluding the input stringearly-return-on-mismatch — The outer loop returns immediately on the first mismatch, delegating to the two skip checks rather than continuing iteration