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• 📝 My LeetCode Solution: 9. Palindrome Number
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LeetCode Submission Results

Explanation of Key Points

Early exits

• Any negative number is not a palindrome (e.g., -121 → false).
• Any number ending in 0 can’t be a palindrome unless it’s 0 itself (e.g., 10 → false).

Reversing half the number

• Iteratively remove the last digit from x and append it to reverted.
• Continue until the remaining part (x) is less than or equal to the reversed half (reverted), meaning we've processed at least half of the digits.

Handling odd/even length

• Even length: The two halves must match exactly (x === reverted).
• Odd length: The reversed half contains the middle digit, so drop it with Math.floor(reverted / 10) before comparing.

This approach satisfies LeetCode’s constraints (−2³¹ ≤ x ≤ 2³¹−1) and avoids any string conversions.

Complexity Analysis

• Time Complexity:
  
  The loop processes roughly half of the digits of x. If x has d digits, it runs ~d/2 times → $$O(d) = O(\log_{10} x)$$.

• Space Complexity:
  
  Only a constant number of integer variables are used → $$O(1)$$.

Code

javascript
/**
 * @param {number} x
 * @return {boolean}
 */
var isPalindrome = function(x) {
    // Negative numbers or numbers ending with 0 (but not 0 itself) cannot be palindrome.
    if (x < 0 || (x % 10 === 0 && x !== 0)) {
        return false;
    }
    let reverted = 0;
    // Reverse digits until the original number is <= reverted reversed half
    while (x > reverted) {
        reverted = reverted * 10 + (x % 10);
        x = Math.floor(x / 10);
    }
    // For even length: x === reverted
    // For odd length: x === Math.floor(reverted / 10)
    return x === reverted || x === Math.floor(reverted / 10);
};