Hard

Super PalindromesC++

Full explanation · Time O(n^0.25 * logn) · Space O(logn)

// Time:  O(n^0.25 * logn)
// Space: O(logn)

class Solution {
public:
    int superpalindromesInRange(string L, string R) {
        const auto K = static_cast<int>(pow(10, (R.length() + 1) * 0.25));
        const int64_t l = stol(L), r = stol(R);
        int result = 0;

        // count odd length
        for (int k = 0; k < K; ++k) {
            const string s = to_string(k), rev_s(s.rbegin(), s.rend());
            int64_t v = stol(s + rev_s.substr(1));
            v *= v;
            if (v > r) {
                break;
            }
            if (v >= l && is_palindrome(v)) {
                ++result;
            }
        }

        // count even length
        for (int k = 0; k < K; ++k) {
            const string s = to_string(k), rev_s(s.rbegin(), s.rend());
            int64_t v = stol(s + rev_s);
            v *= v;
            if (v > r) {
                break;
            }
            if (v >= l && is_palindrome(v)) {
                ++result;
            }
        }

        return result;
    }

private:
    bool is_palindrome(int64_t k) {
        const string s = to_string(k), rev_s(s.rbegin(), s.rend());
        return s == rev_s;
    }
};