Next Smallest Palindrome!

• Time : $O(|s|)$
• Space : $O(1)$

string Solution::solve(string s) {
    int r = s.length() / 2, l = r - (s.length() % 2 == 0);
    int pass = 0;
    while(l >= 0 and pass == 0) {
        if(s[l] == s[r]) l--,r++;
        else if(s[l] > s[r]) pass = 1;
        else pass = -1;
    }
    if(pass != 1) {
        r = s.length() / 2, l = r - (s.length() % 2 == 0);
        int carry = 1;
        while(l >= 0 and carry) {
            int now = s[l] - '0' + carry;
            carry = now / 10;
            now %= 10;
            s[l] = now + '0';
            l--;
        }
        if(carry) s = "1" + s;
    }
    l = 0, r = s.size() - 1;
    while(l < r) {
        s[r] = s[l];
        l++,r--;
    }
    return s;
    
}