Build confidence with an intuition-first walkthrough focused on two pointers fundamentals.
Given a string s, return true if the s can be palindrome after deleting at most one character from it.
Example 1:
Input: s = "aba" Output: true
Example 2:
Input: s = "abca" Output: true Explanation: You could delete the character 'c'.
Example 3:
Input: s = "abc" Output: false
Constraints:
1 <= s.length <= 105s consists of lowercase English letters.Problem summary: Given a string s, return true if the s can be palindrome after deleting at most one character from it.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Two Pointers · Greedy
"aba"
"abca"
"abc"
valid-palindrome)valid-palindrome-iii)valid-palindrome-iv)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #680: Valid Palindrome II
class Solution {
private char[] s;
public boolean validPalindrome(String S) {
this.s = S.toCharArray();
for (int i = 0, j = s.length - 1; i < j; ++i, --j) {
if (s[i] != s[j]) {
return check(i + 1, j) || check(i, j - 1);
}
}
return true;
}
private boolean check(int i, int j) {
for (; i < j; ++i, --j) {
if (s[i] != s[j]) {
return false;
}
}
return true;
}
}
// Accepted solution for LeetCode #680: Valid Palindrome II
func validPalindrome(s string) bool {
check := func(i, j int) bool {
for ; i < j; i, j = i+1, j-1 {
if s[i] != s[j] {
return false
}
}
return true
}
for i, j := 0, len(s)-1; i < j; i, j = i+1, j-1 {
if s[i] != s[j] {
return check(i+1, j) || check(i, j-1)
}
}
return true
}
# Accepted solution for LeetCode #680: Valid Palindrome II
class Solution:
def validPalindrome(self, s: str) -> bool:
def check(i, j):
while i < j:
if s[i] != s[j]:
return False
i, j = i + 1, j - 1
return True
i, j = 0, len(s) - 1
while i < j:
if s[i] != s[j]:
return check(i, j - 1) or check(i + 1, j)
i, j = i + 1, j - 1
return True
// Accepted solution for LeetCode #680: Valid Palindrome II
impl Solution {
pub fn valid_palindrome(s: String) -> bool {
fn is_palindrome(s: &[u8]) -> Option<(usize, usize)> {
let (mut i, mut j) = (0, s.len() - 1);
while i < j {
if s[i] != s[j] {
return Some((i, j));
}
i += 1;
j -= 1;
}
None // is palindrome
}
let s = s.as_bytes().to_vec();
match is_palindrome(&s) {
Some((i, j)) => {
if is_palindrome(&s[i+1..=j]).is_none() {
return true;
}
if is_palindrome(&s[i..=j-1]).is_none() {
return true;
}
}
None => {
return true;
}
}
false
}
}
// Accepted solution for LeetCode #680: Valid Palindrome II
function validPalindrome(s: string): boolean {
const check = (i: number, j: number): boolean => {
for (; i < j; ++i, --j) {
if (s[i] !== s[j]) {
return false;
}
}
return true;
};
for (let i = 0, j = s.length - 1; i < j; ++i, --j) {
if (s[i] !== s[j]) {
return check(i + 1, j) || check(i, j - 1);
}
}
return true;
}
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.
Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.
Review these before coding to avoid predictable interview regressions.
Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.
Usually fails on: A valid pair can be skipped when only one side should move.
Fix: Move exactly one pointer per decision branch based on invariant.
Wrong move: Locally optimal choices may fail globally.
Usually fails on: Counterexamples appear on crafted input orderings.
Fix: Verify with exchange argument or monotonic objective before committing.