Now, let's look at the second approach:

Approach 2: Converting to string and using two pointers

class Solution {
public:
    bool isPalindrome(int x) {
        if (x < 0) {
            return false;
        }
        
        string str = to_string(x);
        int start = 0;
        int end = str.size() - 1;
        
        while (start < end) {
            if (str[start] != str[end]) {
                return false;
            }
            start++;
            end--;
        }
        
        return true;
    }
};

1. Initial check: x is positive, so we proceed.
2. Convert to string: str = "12321"
3. Initialize: start = 0, end = 4
4. Loop iterations:
   • Iteration 1: str[0] == str[4], start = 1, end = 3
   • Iteration 2: str[1] == str[3], start = 2, end = 2
5. Loop ends (start is not < end)
6. Return true

This approach works by converting the number to a string and checking if it reads the same from both ends. It's more intuitive but involves string conversion.

Comparison:

1. Time Complexity: Both are O(log x) or O(n) where n is the number of digits.
2. Space Complexity:
   • Approach 1: O(1), uses constant extra space.
   • Approach 2: O(log x) for string conversion.
3. Readability: Approach 2 is generally more intuitive and easier to understand.
4. Performance: Approach 1 might be slightly faster for very large numbers as it avoids string conversion.

Both approaches are valid and have their merits. Approach 1 is more efficient in terms of space and potentially time for very large numbers, while Approach 2 is more straightforward and easier to understand at a glance. The choice between them might depend on specific requirements or preferences in a given situation.