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Problem solution · C++

Super Palindromes

Super Palindromes: a C++ solution using sliding window or two pointers. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Sliding window or two pointers
Source
walkccc LeetCode Solutions
Length
47 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Sliding window or two pointers

For Super Palindromes, the implementation maintains a moving interval and updates only the information that enters or leaves the window.

  1. Choose the invariant that makes a window valid or useful.
  2. Advance the right boundary and add the new element.
  3. Move the left boundary only as needed while maintaining the invariant and updating the answer.

Code notes

  • 47 lines of C++ from the credited upstream file 906.cpp.
  • The implementation keeps its working state in language-native values and containers.
  • 2 loop blocks detected.

Complexity

Confirm that neither pointer moves backwards; if so, the scan is usually linear apart from the window’s data-structure operations.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeSuper Palindromes · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  int superpalindromesInRange(string left, string right) {    int ans = 0;    const long l = stoll(left);    const long r = stoll(right);     for (long i = sqrt(l); i * i <= r;) {      const long palindrome = nextPalindrome(i);      const long squared = palindrome * palindrome;      if (squared <= r && isPalindrome(squared))        ++ans;      i = palindrome + 1;    }     return ans;  }  private:  long nextPalindrome(int num) {    const string s = to_string(num);    const int n = s.length();    string half = s.substr(0, (n + 1) / 2);    string reversedHalf = reversed(half.substr(0, n / 2));    const long candidate = stoll(half + reversedHalf);    if (candidate >= num)      return candidate;    half = to_string(stoll(half) + 1);    reversedHalf = reversed(half.substr(0, n / 2));    return stoll(half + reversedHalf);  }   string reversed(const string& s) {    return {s.rbegin(), s.rend()};  }   bool isPalindrome(long num) {    const string s = to_string(num);    int l = 0;    int r = s.length() - 1;    while (l < r)      if (s[l++] != s[r--])        return false;    return true;  }};

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