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Problem solution · Java

Super Palindromes

Super Palindromes: a Java 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
45 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

  • 45 lines of Java from the credited upstream file 906.java.
  • 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 · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int superpalindromesInRange(String left, String right) {    int ans = 0;    Long l = Long.valueOf(left);    Long r = Long.valueOf(right);     for (long i = (long) Math.sqrt(l); i * i <= r;) {      long palindrome = nextPalindrome(i);      long squared = palindrome * palindrome;      if (squared <= r && isPalindrome(squared))        ++ans;      i = palindrome + 1;    }     return ans;  }   private long nextPalindrome(long num) {    final String s = String.valueOf(num);    final int n = s.length();     String half = s.substring(0, (n + 1) / 2);    String reversedHalf = new StringBuilder(half.substring(0, n / 2)).reverse().toString();    final long candidate = Long.valueOf(half + reversedHalf);    if (candidate >= num)      return candidate;     half = String.valueOf(Long.valueOf(half) + 1);    reversedHalf = new StringBuilder(half.substring(0, n / 2)).reverse().toString();    return Long.valueOf(half + reversedHalf);  }   private boolean isPalindrome(long num) {    final String s = String.valueOf(num);    int l = 0;    int r = s.length() - 1;     while (l < r)      if (s.charAt(l++) != s.charAt(r--))        return false;     return true;  }}

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