Mead3000 Mead3000 - 1 year ago 72
Java Question

A better algorithm to find the next palindrome of a number string

Firstly here is the problem:

A positive integer is called a palindrome if its representation in the decimal system is the same when read from left to right and from right to left. For a given positive integer K of not more than 1000000 digits, write the value of the smallest palindrome larger than K to output. Numbers are always displayed without leading zeros.

Input: The first line contains integer t, the number of test cases. Integers K are given in the next t lines.

Output: For each K, output the smallest palindrome larger than K.








Secondly here is my code:

// I know it is bad practice to not cater for erroneous input,
// however for the purpose of the execise it is omitted
import java.util.Scanner;
import java.lang.Exception;
import java.math.BigInteger;

public class Main
public static void main(String [] args){
Main instance = new Main(); // create an instance to access non-static
// variables

// Use java.util.Scanner to scan the get the input and initialise the
// variable
Scanner sc=null;

BufferedReader r = new BufferedReader(new InputStreamReader(;

String input = "";

int numberOfTests = 0;

String k; // declare any other variables here

if((input = r.readLine()) != null){
sc = new Scanner(input);
numberOfTests = sc.nextInt();

for (int i = 0; i < numberOfTests; i++){
if((input = r.readLine()) != null){
sc = new Scanner(input);; // initialise the remainder of the variables
} //if
}// for
}// try

catch (Exception e)
}// main

public void palindrome(String number){

StringBuffer theNumber = new StringBuffer(number);
int length = theNumber.length();
int left, right, leftPos, rightPos;
// if incresing a value to more than 9 the value to left (offset) need incrementing
int offset, offsetPos;
boolean offsetUpdated;
// To update the string with new values
String insert;
boolean hasAltered = false;

for(int i = 0; i < length/2; i++){
leftPos = i;
rightPos = (length-1) - i;
offsetPos = rightPos -1; offsetUpdated = false;

// set values at opposite indices and offset
left = Integer.parseInt(String.valueOf(theNumber.charAt(leftPos)));
right = Integer.parseInt(String.valueOf(theNumber.charAt(rightPos)));
offset = Integer.parseInt(String.valueOf(theNumber.charAt(offsetPos)));

if(left != right){
// if r > l then offest needs updating
if(right > left){
// update and replace
right = left;
insert = Integer.toString(right);

theNumber.replace(rightPos, rightPos + 1, insert);

offset++; if (offset == 10) offset = 0;
insert = Integer.toString(offset);

theNumber.replace(offsetPos, offsetPos + 1, insert);
offsetUpdated = true;

// then we need to update the value to left again
while (offset == 0 && offsetUpdated){
offset =
offset++; if (offset == 10) offset = 0;
// replace
insert = Integer.toString(offset);
theNumber.replace(offsetPos, offsetPos + 1, insert);
// finally incase right and offset are the two middle values
left = Integer.parseInt(String.valueOf(theNumber.charAt(leftPos)));
if (right != left){
right = left;
insert = Integer.toString(right);
theNumber.replace(rightPos, rightPos + 1, insert);
}// if r > l
// update and replace
right = left;
insert = Integer.toString(right);
theNumber.replace(rightPos, rightPos + 1, insert);
}// if l != r
}// for i
}// palindrome

Finally my explaination and question.

My code compares either end and then moves in
if left and right are not equal
if right is greater than left
(increasing right past 9 should increase the digit
to its left i.e 09 ---- > 10) and continue to do
so if require as for 89999, increasing the right
most 9 makes the value 90000

before updating my string we check that the right
and left are equal, because in the middle e.g 78849887
we set the 9 --> 4 and increase 4 --> 5, so we must cater for this.

The problem is from an online judge system. My code works for all the test can provide but when I submit it, I get a time limit exceeded error and my answer is not accepted.

Does anyone have any suggestions as to how I can improve my algorithm. While writing this question i thought that instead of my while (offset == 0 && offsetUpdated) loop i could use a boolean to to make sure i increment the offset on my next [i] iteration. Confirmation of my chang or any suggestion would be appreciated, also let me know if i need to make my question clearer.

Answer Source

This seems like a lot of code. Have you tried a very naive approach yet? Checking whether something is a palindrome is actually very simple.

private boolean isPalindrome(int possiblePalindrome) {
    String stringRepresentation = String.valueOf(possiblePalindrome);
    if ( stringRepresentation.equals(stringRepresentation.reverse()) ) {
       return true;

Now that might not be the most performant code, but it gives you a really simple starting point:

private int nextLargestPalindrome(int fromNumber) {
    for ( int i = fromNumber + 1; ; i++ ) {
        if ( isPalindrome( i ) ) {
            return i;

Now if that isn't fast enough you can use it as a reference implementation and work on decreasing the algorithmic complexity.

There should actually be a constant-time (well it is linear on the number of digits of the input) way to find the next largest palindrome. I will give an algorithm that assumes the number is an even number of digits long (but can be extended to an odd number of digits).

  1. Find the decimal representation of the input number ("2133").
  2. Split it into the left half and right half ("21", "33");
  3. Compare the last digit in the left half and the first digit in the right half.
    a. If the right is greater than the left, increment the left and stop. ("22")
    b. If the right is less than the left, stop.
    c. If the right is equal to the left, repeat step 3 with the second-last digit in the left and the second digit in the right (and so on).
  4. Take the left half and append the left half reversed. That's your next largest palindrome. ("2222")

Applied to a more complicated number:

1.    1234567887654322
2.    12345678   87654322
3.    12345678   87654322
             ^   ^         equal
3.    12345678   87654322
            ^     ^        equal
3.    12345678   87654322
           ^       ^       equal
3.    12345678   87654322
          ^         ^      equal
3.    12345678   87654322
         ^           ^     equal
3.    12345678   87654322
        ^             ^    equal
3.    12345678   87654322
       ^               ^   equal
3.    12345678   87654322
      ^                 ^  greater than, so increment the left

3.    12345679

4.    1234567997654321  answer

This seems a bit similar to the algorithm you described, but it starts at the inner digits and moves to the outer.

Recommended from our users: Dynamic Network Monitoring from WhatsUp Gold from IPSwitch. Free Download