Warning: This visualization was AI-generated and may contain inaccuracies. Please verify against the original LeetCode problem (#125) before relying on it.

Valid Palindrome

An interactive visualization of the two-pointer approach to check if a string is a valid palindrome. Watch the left and right pointers converge toward the middle, skipping non-alphanumeric characters.

Two Pointers LeetCode 125 ↗ Easy

The 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. Alphanumeric characters include letters and numbers.

Given a string s, return true if it is a palindrome, or false otherwise.

Example: "A man, a plan, a canal: Panama" → true
Example: "race a car" → false

Interactive Visualization

Input String

Preset

Speed:

Character Array

Current Comparison

Press Step or Play to begin.

Code

fun isPalindrome(s: String): Boolean {    var left = 0; var right = s.length - 1    while (left < right) {        while (left < right && !s[left].isLetterOrDigit()) {            left++  // skip non-alphanumeric        }        while (left < right && !s[right].isLetterOrDigit()) {            right--  // skip non-alphanumeric        }        if (s[left].lowercaseChar() != s[right].lowercaseChar()) {            return false        }        left++        right--    }    return true}

def isPalindrome(s):    left, right = 0, len(s) - 1    while left < right:        while left < right and not s[left].isalnum():            left += 1  # skip non-alphanumeric        while left < right and not s[right].isalnum():            right -= 1  # skip non-alphanumeric        if s[left].lower() != s[right].lower():            return False        left += 1        right -= 1    return True

Status & Log

Press Step or Play to begin.

How It Works

Key Insight

A palindrome reads the same forwards and backwards. Instead of creating a new cleaned string (which uses O(n) extra space), we use two pointers starting from opposite ends and moving inward. We skip any non-alphanumeric characters and compare the remaining characters case-insensitively. If every pair matches, the string is a palindrome.

The Algorithm

Initialize: left = 0 (start of string), right = len(s) - 1 (end of string)
While left < right:
- Skip non-alphanumeric characters by advancing left forward
- Skip non-alphanumeric characters by moving right backward
- Compare s[left].lower() with s[right].lower()
- If they differ, return False immediately
- If they match, move both pointers inward: left += 1, right -= 1
If all pairs matched: return True

Complexity

Time O(n)

Space O(1)

Each character is visited at most once by the left pointer and at most once by the right pointer, giving O(n) time. We use only a constant amount of extra space for the two pointer variables.