Question

Problem statement: For a given positive number, I have to find out the immediately next palindrome. Eg:

For 808, output:818
2133, output:2222

I want to know if my code is efficient at all, and how efficient is it? Is this a good way to solve the problem?

Logic explanation: I’ve set i to the leftmost position of the number, j to the rightmost position an I’m basically comparing the 2 numbers. I assign num[j]=num[i] always, and keep track if the number becomes greater than the original value, or lesser, or equal. In the end,that is: j-i==1 or j==i, depending on number of digits of the number being even or odd, I see if the number became greater or not, taking a decision accordingly.

EDIT: The number can be upto 100,000 digits long!..That was part of the problem statement,so I’m trying to avoid brute force methods.

int LeftNineIndex = 0, RightNineIndex = 0;
bool NumberLesser = false, NumberGreater = false;
string number = Console.ReadLine();
char[] num = number.ToCharArray();
int i, j, x, y;
for (i = 0, j = num.Length - 1; i <= j; i++, j--)
{
      char m;
      Int32.TryParse(num[i].ToString(),out x);
      Int32.TryParse(num[j].ToString(), out y);
      if (x > y)
      {
           NumberGreater = true;
           NumberLesser = false;
      }
      else if (x < y)
      {
           if (j - i == 1)
           {
                NumberGreater = true;
                NumberLesser = false;
                x = x + 1;
                Char.TryParse(x.ToString(), out m);
                num[i] = m;
           }
           else
           {
                NumberGreater = false;
                NumberLesser = true;
           }             
     }

     if ((j == i && NumberGreater == false) || (j - i == 1 && x == y &&  NumberGreater == false))
     {
           if (x != 9)  // if the number is 9, then i can't add 1 to it
           {
                x = x + 1;
                Char.TryParse(x.ToString(), out m);
                num[i] = m;
           }
           else
           {
                if (num.Length != 1)
                {
                    Int32.TryParse(num[LeftNineIndex].ToString(), out x);
                    Int32.TryParse(num[RightNineIndex].ToString(), out y);
                    x = x + 1;
                    Char.TryParse(x.ToString(), out m);
                    num[LeftNineIndex] = m;
                    num[RightNineIndex] = m;
                }
                else
                {
                    // user has entered just '9', in which case I've hard-coded
                    Console.WriteLine("11");
                }
           }
     }
     num[j] = num[i];
     if (x != 9)  //gives us the index of the number closest to the middle, which is not 9
     {
          LeftNineIndex = i;
          RightNineIndex = j;
     }
}
Console.WriteLine(num);

Answer 1

It’s relatively simple to find the next palindrome in constant time:

• Split the input in two halves (first half larger if length is odd)
• Now for the next palindrome there are two candidates:
  1. Keep the first half, and fix the second half
  2. Increment the first half by 1, and fix the second half
• Choose the first candidate if it is larger than the input, choose the second candidate otherwise.

The BigInteger type is useful for implementing this:

This approach has a cost linear in the length of the input, i.e. it’s logarithmic in the size of the number.

public static BigInteger NextPalindrome(BigInteger input)
{
    string firstHalf=input.ToString().Substring(0,(input.ToString().Length+1)/2);
    string incrementedFirstHalf=(BigInteger.Parse(firstHalf)+1).ToString();
    var candidates=new List<string>();
    candidates.Add(firstHalf+new String(firstHalf.Reverse().ToArray()));
    candidates.Add(firstHalf+new String(firstHalf.Reverse().Skip(1).ToArray()));
    candidates.Add(incrementedFirstHalf+new String(incrementedFirstHalf.Reverse().ToArray()));
    candidates.Add(incrementedFirstHalf+new String(incrementedFirstHalf.Reverse().Skip(1).ToArray()));
    candidates.Add("1"+new String('0',input.ToString().Length-1)+"1");
    return candidates.Select(s=>BigInteger.Parse(s))
              .Where(i=>i>input)
              .OrderBy(i=>i)
              .First();
}

Tested to work with all positive integers below 100000 by comparing with the native implementation.

The fifth case is easy to miss. If the number consists of all 9s, incrementing the first half changes the length, and requires extra handling if the current length is odd (9, 999,…).