A “Palindrome Number” is a number that remains the same when its digits are reversed.

For example, 121 and 454 are palindrome numbers, but 123 and 789 are not. The problem is to determine whether a given integer is a palindrome or not?

To solve this problem in C++, we can follow the steps below:

  1. Convert the integer to a string.
  2. Compare the characters from both ends of the string, moving towards the middle, to check if they are the same.

Here’s an optimized C++ solution for the “Palindrome Number” problem:

#include <iostream>
using namespace std;

bool isPalindrome(int x) {
    if (x < 0) {
        return false; // Negative numbers are not palindromes
    }

    string numStr = to_string(x);
    int left = 0;
    int right = numStr.length() - 1;

    while (left < right) {
        if (numStr[left] != numStr[right]) {
            return false;
        }
        left++;
        right--;
    }

    return true;
}

int main() {
    int num;
    cout << "Enter an integer: ";
    cin >> num;

    if (isPalindrome(num)) {
        cout << num << " is a palindrome." << endl;
    } else {
        cout << num << " is not a palindrome." << endl;
    }

    return 0;
}

This solution first handles negative numbers by returning false since negative numbers cannot be palindromes. Then, it converts the given integer to a string and uses two pointers (left and right) to compare characters from both ends of the string. If the characters at left and right indices are not the same, it returns false. Otherwise, it continues moving towards the middle of the string. If the loop completes without finding any mismatch, it means the number is a palindrome, and the function returns true.

This approach has a time complexity of O(log N), where N is the number of digits in the input integer, and a space complexity of O(log N) due to the conversion of the integer to a string. However, the space complexity can be considered constant in practice, as the length of the string representation is limited for typical integer values.