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javaoptimizationpalindrome

USACO Calf Flac Palindrone Times Out


This is the problem statement: Ignore punctuation, whitespace, numbers, and case when testing for palindromes, but keep these extra characters around so that you can print them out as the answer; just consider the letters A-Z' anda-z'.

Find the largest palindrome in a string no more than 20,000 characters long. The largest palindrome is guaranteed to be at most 2,000 characters long before whitespace and punctuation are removed.

The first line of the output should be the length of the longest palindrome found. The next line or lines should be the actual text of the palindrome (without any surrounding white space or punctuation but with all other characters) printed on a line (or more than one line if newlines are included in the palindromic text). If there are multiple palindromes of longest length, output the one that appears first.

I'm trying to iterate through the String with only letters, and testing that i (odd length) or i and i - 1 (even length) is the middle of a palidrone. My code works, but it's too slow, and times out on test case 8 of 10. Currently, the test takes 1.9 seconds, and the max limit is 1 second.

My code

import java.io.*;
import java.util.*;

class calfflac { 
  public static void main (String [] args) throws IOException {
    BufferedReader f = new BufferedReader(new FileReader("calfflac.in"));
    PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter("calfflac.out")));
    String line = "";
    while(true) {
        String l = f.readLine();
        if(l == null) break;
        line += l + "\n";
    }
    String temp = line;
    line = line.toLowerCase();
    String rem = "";
    ArrayList<Integer> map = new ArrayList<Integer>();
    /* Mapping the index of the letter in the String with parsed chars
     *  to the index of the letter in the original String */
    for(int i = 0; i < line.length(); i++) {
        if(Character.isLetter(line.charAt(i))) {
            map.add(i);
            rem += line.substring(i, i + 1);
        }
    }
    int longest = 0; // longest palindrome length
    int left = 0; // first char index of the palindrome
    int right = 0; // last char index of the palindrome
    for(int i = 1; i < rem.length() - longest / 2; i++) {
        int[] n = getPali(i, i, rem, longest / 2); // odd length palidrome
        if(n[1] - n[0] > longest && n[0] != i) {
            longest = n[1] - n[0] + 1;
            left = n[0];
            right = n[1];
        }
        n = getPali(i - 1, i, rem, longest / 2); // even length palindrome
        if(n[1] - n[0] > longest && n[0] != i - 1) {
            longest = n[1] - n[0] + 1;
            left = n[0];
            right = n[1];
        }
    }
    out.println(longest);
    // getting the indices of the first and last char in the original substring and printing it.
    out.println(temp.substring(map.get(left), map.get(right) + 1));
    out.close();
    f.close();
    System.exit(0);
  }

  // checks if a given String is a palindrome
  public static boolean isPali(String str) {
      for(int i = 0; i < str.length() / 2 + 1; i++) {
          if(str.charAt(i) != str.charAt(str.length() - 1 - i)) {
              return false;
          }
      }
      return true;
  }

  /*
   * returns the indices of the first and last chars of the largest palindrome with left, right as the middle
   * right == left is a palindrome of odd length
   * right - left == 1 is a palindrome of even length
   */
  public static int[] getPali(int left, int right, String str, int longest) {
      if(left - longest < 0 || right + longest >= str.length()) {
          return new int[]{left, right};
      }
      // checks if the String is a palidrome. I think I can somehow delete this check, but not sure how
      if(!isPali(str.substring(left - longest, right + longest + 1))) {
          return new int[]{left, right};
      }

      left = left - longest;
      right = right + longest;
      for(int i = 0; i < str.length() / 2; i++) {
          if(left < 0 || right > str.length() - 1) return new int[]{left + 1, right - 1};
          if(str.charAt(left) == str.charAt(right)) { // compares the next two chars. Checks inside out
              left--;
              right++;
          } else {
              break;
          }
      }
      return new int[]{left + 1, right - 1};
  }


}

Fails this check:

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I think I can make it faster by somehow doing the even palindrome and odd palindrome at the same time, but I'm not sure how. I'm also thinking that it might be possible to not have to test the inside of the palindrome if it's shorter then the longest palindrome I've found so far. Could someone give me a hint on to how optimize or make my code faster?


Solution

  • I think you are over complicating the problem, I'll try to help but my java is a little rusty, so i'll write the code in a very lose form, i'll explain my solution in an edit.

    int maxPalindrom(String x){
        int max = 0;
        int current;
        for(int i=0; i<x.length - 1; i++){
            current = 1;
            for(int j = 1; j < min(x.length - i, i); j ++){
                if(x[i-j] == x[i+j])
                    current += 2;
                else
                    break;
            }
            if (current > max) current = max;
            current = 0;
            for(int j = 0; j < min(x.length - i, i); j ++){
                if(x[i - j] == x[i + 1 - j])
                    current += 2;
                else
                    break;
            }
            if (current > max) current = max;
        }
        return max;
    }
    

    update

    a palindrome could be either in the form of aaxaa or axxa so i iterate on the string to and i move i so it would be at the x on the first inner loop and on the first x in the second inner loop, I update current until i find a case that will break the palindrome. I update max if current has a higher value. you can further optimise this code by considering the max value in the outer loop

    I hope this was helpful, if you have further questions don't hesitate to ask.