Problem Statement

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 a palindrome, or false otherwise.

LeetCode

• https://leetcode.com/problems/valid-palindrome/description/

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
Explanation: "amanaplanacanalpanama" is a palindrome.

Example 2:

Input: s = "race a car"
Output: false
Explanation: "raceacar" is not a palindrome.

Example 3:

Input: s = " "
Output: true
Explanation: s is an empty string "" after removing non-alphanumeric characters.
Since an empty string reads the same forward and backward, it is a palindrome.

Different Approaches

1️⃣ Two Pointers

🧠 Idea:

• Use two pointers:
  • one from the start and one from the end.
  • Skip non-alphanumerics, compare lowercase letters.

Code:

bool isPalindrome(string s) {
    int left = 0, right = s.size() - 1;
    while (left < right) {
        // Skip non-alphanumerics
        while (left < right && !isalnum(s[left])) left++;
        while (left < right && !isalnum(s[right])) right--;

        // Compare lowercased characters
        if (tolower(s[left]) != tolower(s[right])) return false;

        left++;
        right--;
    }
    return true;
}

Complexity Analysis:

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

2️⃣ Clean and Reverse

🧠 Idea:

• Clean the string (only alphanumeric + lowercase).
• Compare it with its reverse.

Code:

bool isPalindrome(string s) {
    string clean = "";
    for (char c : s) {
        if (isalnum(c)) clean += tolower(c);
    }

    string rev = clean;
    reverse(rev.begin(), rev.end());
    return clean == rev;
}

Complexity Analysis:

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

3️⃣ Recursive

🧠 Idea:

1. Use a helper function that takes two indices left and right.
2. Skip non-alphanumeric characters.
3. Compare lowercase characters.
4. Recur inward.

Code:

class Solution {
public:
    bool isPalindrome(string& s, int left, int right) {
        // Base case: pointers crossed
        if (left >= right) return true;

        // Skip non-alphanumerics
        if (!isalnum(s[left])) return isPalindrome(s, left + 1, right);
        if (!isalnum(s[right])) return isPalindrome(s, left, right - 1);

        // Compare lowercase characters
        if (tolower(s[left]) != tolower(s[right])) return false;

        // Recur inward
        return isPalindrome(s, left + 1, right - 1);
    }

    bool isPalindrome(string s) {
        return isPalindrome(s, 0, s.size() - 1);
    }
};

Complexity Analysis:

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