The challenge
Number is a palindrome if it is equal to the number with digits in reversed order. For example, 5
, 44
, 171
, 4884
are palindromes, and 43
, 194
, 4773
are not.
Write a function which takes a positive integer and returns the number of special steps needed to obtain a palindrome. The special step is: “reverse the digits, and add to the original number”. If the resulting number is not a palindrome, repeat the procedure with the sum until the resulting number is a palindrome.
If the input number is already a palindrome, the number of steps is 0
.
All inputs are guaranteed to have a final palindrome which does not overflow long
.
Example
For example, start with 87
:
87 + 78 = 165 - step 1, not a palindrome 165 + 561 = 726 - step 2, not a palindrome 726 + 627 = 1353 - step 3, not a palindrome 1353 + 3531 = 4884 - step 4, palindrome!
4884
is a palindrome and we needed 4
steps to obtain it, so answer for 87
is 4
.
Additional info
Some interesting information on the problem can be found in this Wikipedia article on Lychrel numbers.
The solution in Java code
Option 1:
public class Palindromes { public static int palindromeChainLength (long n) { String ns = "" + n, nrs = "" + new StringBuilder(ns).reverse(); return ns.equals(nrs) ? 0 : 1 + palindromeChainLength(n + Long.valueOf(nrs)); } }
Option 2:
public class Palindromes { public static int palindromeChainLength(long n) { int step = 0; while (n != reverseNumber(n)) { n = n + reverseNumber(n); step++; } return step; } private static long reverseNumber(long n) { char[] number = String.valueOf(n).toCharArray(); String reverseNumber = ""; for (int count = number.length - 1; count >= 0; count--) { reverseNumber = reverseNumber + number[count]; } return Long.parseLong(reverseNumber); } }
Option 3:
public class Palindromes { public static int palindromeChainLength (long n) { String s = String.valueOf(n); String r = new StringBuilder(String.valueOf(n)).reverse().toString(); if (s.equals(r)) return 0; return 1 + palindromeChainLength(n + Long.parseLong(r)); } }
Test cases to validate our solution
import org.junit.Test; import static org.junit.Assert.*; public class PalindromesTest { @Test public void testPalindrome() { assertEquals(0, Palindromes.palindromeChainLength(1)); assertEquals(0, Palindromes.palindromeChainLength(88)); assertEquals(0, Palindromes.palindromeChainLength(393)); } @Test public void testNonPalindrome() { assertEquals(1, Palindromes.palindromeChainLength(10)); assertEquals(1, Palindromes.palindromeChainLength(134)); assertEquals(4, Palindromes.palindromeChainLength(87)); assertEquals(7, Palindromes.palindromeChainLength(2897)); assertEquals(24, Palindromes.palindromeChainLength(89)); } }