Problem:
Score : 400 points

Problem Statement

A positive integer X is said to be a lunlun number if and only if the following condition is satisfied:

In the base ten representation of X (without leading zeros), for every pair of two adjacent digits, the absolute difference of those digits is at most 1.

For example, 1234, 1, and 334 are lunlun numbers, while none of 31415, 119, or 13579 is.

You are given a positive integer K. Find the K-th smallest lunlun number.

Constraints

1 \leq K \leq 10^5

All values in input are integers.

Input

Input is given from Standard Input in the following format:

K

Output

Print the answer.

Sample Input 1

15

Sample Output 1

23

We will list the 15 smallest lunlun numbers in ascending order:

1,
2,
3,
4,
5,
6,
7,
8,
9,
10,
11,
12,
21,
22,
23.

Thus, the answer is 23.

Sample Input 2

1

Sample Output 2

1

Sample Input 3

13

Sample Output 3

21

Sample Input 4

100000

Sample Output 4

3234566667

Note that the answer may not fit into the 32-bit signed integer type.

Solution in Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.*;

public class Main {
    public static void main(String[] args) throws Exception {
        // Your code here! 
        Scanner sc = new Scanner(System.in);
        long n = sc.nextLong();
        
        // que: 七五三数(候補) (全ての桁が7 or 5 or 3)
        Queue<Long> que = new ArrayDeque<Long>();
        
        // 3, 5, 7からBFSスタート
        // (後ろの3項は、1なら3,5,7が既に登場)
        for (long i = 1; i <= 9; i++) {
            que.add(i);
        }
        
        // count: n以下の七五三数の個数
        // -> 全ての桁が7 or 5 or 3かつ、7,5,3全て登場する数
        // ex. 753, 3577735, 533333337
        int count = 0;
        while (!que.isEmpty()) {
            long num = que.poll();
            // System.out.println(num);
            count++;
            if (count == n) {
                System.out.println(num);
                return;
            }
            
            long x10 = num * 10;
            long last = num % 10;
            
            // 次の桁のパターンをキューに入れる x10+3, x10+5, x10+7
            if (last >= 1) que.add(x10 + last - 1);
            que.add(x10 + last);
            if (last < 9) que.add(x10 + last + 1);
        }
    }
}