Problem:
Score : 400 points

Problem Statement

Given is a positive integer N.

Find the number of pairs (A, B) of positive integers not greater than N that satisfy the following condition:

When A and B are written in base ten without leading zeros, the last digit of A is equal to the first digit of B, and the first digit of A is equal to the last digit of B.

Constraints

1 \leq N \leq 2 \times 10^5

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N

Output

Print the answer.

Sample Input 1

25

Sample Output 1

17

The following 17 pairs satisfy the condition: (1,1), (1,11), (2,2), (2,22), (3,3), (4,4), (5,5), (6,6), (7,7), (8,8), (9,9), (11,1), (11,11), (12,21), (21,12), (22,2), and (22,22).

Sample Input 2

1

Sample Output 2

1

Sample Input 3

100

Sample Output 3

108

Sample Input 4

2020

Sample Output 4

40812

Sample Input 5

200000

Sample Output 5

400000008

Solution in Scala (2.13.1) (judge verdict: Accepted)

object Main {
  def main(args: Array[String]): Unit = {
    val num = io.StdIn.readInt()
    var first_last_num_pair: Map[( Int, Int ), Int ] = Map.empty
    (1 to num).foreach { i =>
      val pair = pairFirstAndLast( i )
      if( first_last_num_pair.contains( pair ) ) first_last_num_pair = first_last_num_pair + ( pair -> ( first_last_num_pair( pair ) + 1 ) )
      else                                       first_last_num_pair = first_last_num_pair + ( pair -> 1 )
    }

    var ans = 0
    ( 1 to num ).foreach{ i =>
      val pair = pairFirstAndLast( i )
      val change_pair = ( pair._2, pair._1 )
      if( first_last_num_pair.contains( change_pair ) ) ans += first_last_num_pair( change_pair )
    }
    println( ans )
  }

    def pairFirstAndLast( num: Int ): ( Int, Int ) = {
      val num_str   = num.toString
      val first_num = num_str(0).toInt - 48
      val last_num  = num_str( num_str.length - 1 ).toInt - 48

      return ( first_num, last_num )
    }
}