17. Valid Palindrome

Q: Why not just use String.prototype.normalize?

normalize handles Unicode equivalence but not alphanumeric filtering. The interview wants you to demonstrate the explicit two-pointer logic.

Q: What if the string has Unicode like 'café'?

The regex /[a-z0-9]/i excludes Unicode letters. For real Unicode support, use Intl.Collator or normalize + a broader regex. Mention this as a follow-up consideration.

easyAsked at Salesforce

Determine if a string is a palindrome considering only alphanumeric characters and ignoring case. Salesforce uses this to test two-pointer string manipulation with edge-case handling.

By Alex Chen, Founder, InterviewChamp.AI · Last verified 2026-05-20

Source citations

Public interview reports confirming this problem appears in Salesforce loops.

Glassdoor (2026-Q1)— Common warmup on Salesforce frontend phone screens.
LeetCode Discuss (2025-09)— Used to test string normalization fluency.

Problem

A phrase is a palindrome if, after converting all uppercase letters into lowercase letters and removing all non-alphanumeric characters, it reads the same forward and backward. Given a string s, return true if it is a palindrome, or false otherwise.

Constraints

1 <= s.length <= 2 * 10^5
s consists only of printable ASCII characters.

Examples

Example 1

Input

s = "A man, a plan, a canal: Panama"

Output

true

Example 2

Input

s = "race a car"

Output

false

Example 3

Input

s = " "

Output

true

Approaches

1. Filter then reverse-compare

Filter alphanumerics, lowercase, compare to its reverse.

Time: O(n)
Space: O(n)

function isPalindrome(s) {
  const cleaned = s.toLowerCase().replace(/[^a-z0-9]/g, '');
  return cleaned === [...cleaned].reverse().join('');
}

Tradeoff: Easy to read but allocates a new string. Salesforce will push for in-place two-pointer.

2. Two-pointer in place

Pointers from both ends. Skip non-alphanumerics. Compare lowercased values. Move inward.

Time: O(n)
Space: O(1)

function isPalindrome(s) {
  const isAlnum = (c) => /[a-z0-9]/i.test(c);
  let l = 0, r = s.length - 1;
  while (l < r) {
    while (l < r && !isAlnum(s[l])) l++;
    while (l < r && !isAlnum(s[r])) r--;
    if (s[l].toLowerCase() !== s[r].toLowerCase()) return false;
    l++; r--;
  }
  return true;
}

Tradeoff: O(1) extra space. The inner while-loops handle arbitrary-length runs of non-alphanumerics.

Salesforce-specific tips

Salesforce values fluency with string normalization (lowercase, alphanumeric filter) because their SOQL identifier matching uses similar rules. Bonus signal: mention that the two-pointer version short-circuits on the first mismatch, which is faster than allocating a full cleaned string and reversing.

Common mistakes

Forgetting l < r inside the inner while-loops — pointers cross when the string is all non-alphanumeric.
Comparing without lowercasing — fails 'A man, a plan...' case.
Using ASCII range checks instead of regex — works but is brittle to non-ASCII inputs.

Follow-up questions

An interviewer at Salesforce may pivot to one of these next:

Valid Palindrome II (LC 680) — allow one deletion.
Longest Palindromic Substring (LC 5).
Palindrome Linked List (LC 234).

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Output

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FAQ

Why not just use String.prototype.normalize?

normalize handles Unicode equivalence but not alphanumeric filtering. The interview wants you to demonstrate the explicit two-pointer logic.

What if the string has Unicode like 'café'?

The regex /[a-z0-9]/i excludes Unicode letters. For real Unicode support, use Intl.Collator or normalize + a broader regex. Mention this as a follow-up consideration.

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