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Valid Palindrome II LC #680 Easy Greedy Two Pointers
Problem

Given a string s, return true if the s can be a palindrome after deleting at most one character from it.

Example 1
Input: s = "abca"
Output: true
Explanation: Delete 'c' to get "aba".
Example 2
Input: s = "abc"
Output: false
Explanation: Neither skip-l nor skip-r yields a palindrome.
Constraints: 1 ≤ s.length ≤ 10⁵ | s consists of lowercase English letters
Try Examples
Custom:
Approach
Greedy two pointers — on mismatch, try skip-l OR skip-r
Scan inward while chars match. At first mismatch, use your one deletion: check if s[l+1..r] or s[l..r-1] is a palindrome via isPalin. Either works → true. O(n) time, O(1) space.
String Strip
deleted char
remaining substring
— No deletion yet — scan with l and r
Variables
l
0
r
—
skip
—
branch
—
result
—
Step Logic
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Algorithm
1
Init l = 0, r = n−1
2
While l < r: if s[l] == s[r], move both inward
3
On mismatch: try isPalin(s, l+1, r) OR isPalin(s, l, r−1)
4
Either substring is palindrome → return true
5
Neither works → false; loop ends with no mismatch → true
Time
O(n)
Space
O(1)
Why This Works

Greedy: only act at the first mismatch. Before that point, chars already matched — no deletion needed there. With exactly one deletion allowed, only two choices exist: drop the left char or the right char. Run isPalin on each remaining substring; if either is a palindrome, the answer is true. Both must fail for false.