Palindrome Sentence

Check whether a sentence reads the same forwards and backwards once you ignore case, spaces and punctuation. Practises the two-pointer technique on strings.

By Deepak KumarNov 12, 2026, 09:22 PM ISTTests checked by DeepakJavaScript or Python

Given a string s, return true if it is a palindrome after converting letters to lowercase and removing every character that isn't a letter or a digit. Otherwise return false.

Only the ASCII letters a-z, A-Z and digits 0-9 count.

Examples

Example 1

Input: s = "Never odd or even."
Output: true
Why: "neveroddoreven" reads the same both ways

Example 2

Input: s = "Node, not done"
Output: false

Constraints

0 <= s.length <= 2 * 10^5
s contains printable ASCII characters

Hints

Hint 1

Building a cleaned copy of the string works. Can you check it without building a copy?

Hint 2

Use one pointer from each end; skip characters that don't count, compare the rest in lowercase.

Solution

Show the solution (try it yourself first)

Approach: two pointers

Start one pointer at each end of the string. Move each pointer past characters that aren't letters or digits, then compare the two characters in lowercase. If they differ, it isn't a palindrome; otherwise move both pointers inward. This avoids building a cleaned copy, so it uses constant extra space. Watch the trap in "0P": digits count too, and "0" is not "p". A string with no letters or digits at all, like ".,!", is a palindrome because there is nothing to mismatch.

function isPalindromeSentence(s) {
  const ok = (c) => /[a-z0-9]/i.test(c);
  let i = 0, j = s.length - 1;
  while (i < j) {
    if (!ok(s[i])) { i++; continue; }
    if (!ok(s[j])) { j--; continue; }
    if (s[i].toLowerCase() !== s[j].toLowerCase()) return false;
    i++; j--;
  }
  return true;
}

Complexity

Time O(n), each character is visited at most once. Space O(1). The clean-then-reverse version is shorter but uses O(n) extra memory.