Palindromes using Scala

Question

I came across this problem from CodeChef. The problem states the following:

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.

I can define a isPalindrome method as follows:

def isPalindrome(someNumber:String):Boolean = someNumber.reverse.mkString == someNumber

The problem that I am facing is how do I loop from the initial given number and break and return the first palindrome when the integer satisfies the isPalindrome method? Also, is there a better(efficient) way to write the isPalindrome method?

It will be great to get some guidance here

Answer 1

If you have a number like 123xxx you know, that either xxx has to be below 321 - then the next palindrom is 123321.

Or xxx is above, then the 3 can't be kept, and 124421 has to be the next one.

Here is some code without guarantees, not very elegant, but the case of (multiple) Nines in the middle is a bit hairy (19992):

object Palindrome extends App {

def nextPalindrome (inNumber: String): String = {
  val len = inNumber.length ()
  if (len == 1 && inNumber (0) != '9') 
    "" + (inNumber.toInt + 1) else {
    val head = inNumber.substring (0, len/2)
    val tail = inNumber.reverse.substring (0, len/2)
    val h = if (head.length > 0) BigInt (head) else BigInt (0)
    val t = if (tail.length > 0) BigInt (tail) else BigInt (0)

    if (t < h) {
      if (len % 2 == 0) head + (head.reverse)
      else inNumber.substring (0, len/2 + 1) + (head.reverse)
    } else {
     if (len % 2 == 1) {
       val s2 = inNumber.substring (0, len/2 + 1) // 4=> 4
       val h2 = BigInt (s2) + 1  // 5 
       nextPalindrome (h2 + (List.fill (len/2) ('0').mkString)) // 5 + "" 
     } else {
       val h = BigInt (head) + 1
       h.toString + (h.toString.reverse)
     }
    }
  }
}

def check (in: String, expected: String) = {
  if (nextPalindrome (in) == expected) 
    println ("ok: " + in) else 
    println (" - fail: " + nextPalindrome (in) + " != " + expected + " for: " + in)
}
//
val nums = List (("12345", "12421"), // f
  ("123456", "124421"), 
  ("54321", "54345"), 
  ("654321", "654456"), 
  ("19992", "20002"),
  ("29991", "29992"),
  ("999", "1001"),
  ("31", "33"),
  ("13", "22"),
  ("9", "11"),
  ("99", "101"),
  ("131", "141"),
  ("3", "4")
)
nums.foreach (n => check (n._1, n._2))
println (nextPalindrome ("123456678901234564579898989891254392051039410809512345667890123456457989898989125439205103941080951234566789012345645798989898912543920510394108095"))

}

I guess it will handle the case of a one-million-digit-Int too.

Answer 2

Doing reverse is not the greatest idea. It's better to start at the beginning and end of the string and iterate and compare element by element. You're wasting time copying the entire String and reversing it even in cases where the first and last element don't match. On something with a million digits, that's going to be a huge waste.

This is a few orders of magnitude faster than reverse for bigger numbers:

def isPalindrome2(someNumber:String):Boolean = {
  val len = someNumber.length;
  for(i <- 0 until len/2) {
    if(someNumber(i) != someNumber(len-i-1)) return false; 
  }
  return true;
}

There's probably even a faster method, based on mirroring the first half of the string. I'll see if I can get that now...

update So this should find the next palindrome in almost constant time. No loops. I just sort of scratched it out, so I'm sure it can be cleaned up.

def nextPalindrome(someNumber:String):String = {
  val len = someNumber.length;
  if(len==1) return "11";
  val half = scala.math.floor(len/2).toInt;
  var firstHalf = someNumber.substring(0,half);
  var secondHalf = if(len % 2 == 1) {
    someNumber.substring(half+1,len);
  } else {
    someNumber.substring(half,len);
  }

  if(BigInt(secondHalf) > BigInt(firstHalf.reverse)) {
    if(len % 2 == 1) {
      firstHalf += someNumber.substring(half, half+1);
      firstHalf = (BigInt(firstHalf)+1).toString;
      firstHalf + firstHalf.substring(0,firstHalf.length-1).reverse
    } else {
      firstHalf = (BigInt(firstHalf)+1).toString;
      firstHalf + firstHalf.reverse;
    }
  } else {
    if(len % 2 == 1) {
      firstHalf + someNumber.substring(half,half+1) + firstHalf.reverse;
    } else {
      firstHalf + firstHalf.reverse;
    }
  }
}