Hard
Super Palindromes — C++
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;
}
};