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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.Scanner;

public class Main {
    private static final int MOD = 1_000_000_007;
    private static final String YES = "Yes";
    private static final String NO = "No";

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int K = sc.nextInt();
        System.out.println(solveDFSNonIter(K));
    }

    private static String solve(int K) {
        if (K <= 9) {
            return String.valueOf(K);
        }

        int total = 9;
        List<String>[] ans = new ArrayList[11];
        List<String>[] leading0 = new ArrayList[11];
        for (int i = 0; i < 11; i++) {
            ans[i] = new ArrayList<>();
            leading0[i] = new ArrayList<>();
        }
        for (long i = 1; i <= 9; i++) {
            ans[1].add(String.valueOf(i));
        }
        leading0[1].add(String.valueOf(0));

        for (int i = 1; i <= 9; i++) {
            for (String val : leading0[1]) {
                if (Math.abs(val.charAt(0) - '0' - i) <= 1) {
                    ans[2].add(i + val);
                }
            }
            for (String val : ans[1]) {
                if (Math.abs(val.charAt(0) - '0' - i) <= 1) {
                    ans[2].add(i + val);
                }
            }
        }
        if (K <= total + ans[2].size()) {
            /*
            for (int i=1; i<=10; i++) {
                System.err.print(i + "(" + ans[i].size() + "):");
                for (String val : ans[i]) {
                    System.err.print(" " + val);
                }
                System.err.println();
            }
             */

            return ans[2].get(K - total - 1);
        }
        total += ans[2].size();
        leading0[2].add("00");
        leading0[2].add("01");

        for (int k = 3; k <= 10; k++) {
            for (int i = 1; i <= 9; i++) {
                for (String val : leading0[k - 1]) {
                    if (Math.abs(val.charAt(0) - '0' - i) <= 1) {
                        ans[k].add(i + val);
                    }
                }

                for (String val : ans[k - 1]) {
                    if (Math.abs(val.charAt(0) - '0' - i) <= 1) {
                        ans[k].add(i + val);
                    }
                }
            }
            Collections.sort(ans[k]);

            if (K <= total + ans[k].size()) {
                /*
                for (int i=1; i<=10; i++) {
                    System.err.print(i + "(" + ans[i].size() + "):");
                    for (String val : ans[i]) {
                        System.err.print(" " + val);
                    }
                    System.err.println();
                }
                 */

                return ans[k].get(K - total - 1);
            }
            total += ans[k].size();
            for (String val : leading0[k-1]) {
                leading0[k].add("0" + val);
            }
            for (String val : ans[k-1]) {
                if (Math.abs(val.charAt(0) - '0') <= 1) {
                    leading0[k].add("0" + val);
                }
            }
        }

        return "-1";
    }

    /**
     * https://drken1215.hatenablog.com/entry/2020/04/05/150900 解法1-1
     * 最大で10桁までとわかるのでDFSで列挙する
     * @param K
     * @return
     */
    private static String solveDFS(int K) {
        List<Long> list = new ArrayList<>();

        for (int v=1; v<10; v++) {
            solveDFSIter(1, v, list);
        }

        Collections.sort(list);

        return String.valueOf(list.get(K-1));
    }

    private static void solveDFSIter(int d, long val, List<Long> list) {
        list.add(val);

        if (d == 10) return;

        for (int j=-1; j<=1; j++) {
            int add = (int)val%10 + j;
            if (add >= 0 && add <= 9) {
                solveDFSIter(d+1, val*10+add, list);
            }
        }
    }

    /**
     * https://drken1215.hatenablog.com/entry/2020/04/05/150900 解法1-2
     * DFSで列挙する際にd桁のものを順にすべて求める
     * @param K
     * @return
     */
    private static String solveDFSNonIter(int K) {
        List<Long> all = new ArrayList<>();

        List<Long> cur = new ArrayList<>();
        for (int v=1; v<10; v++) {
            cur.add((long)v);
            all.add((long)v);
        }

        for (int d=1; d<10; d++) {
            cur = calcNext(cur);
            all.addAll(cur);
        }

        return String.valueOf(all.get(K-1));
    }

    private static List<Long> calcNext(List<Long> cur) {
        List<Long> res = new ArrayList<>();

        for (long val : cur) {
            for (int j=-1; j<=1; j++) {
                int add = (int)(val %10) + j;
                if (add >= 0 && add <= 9) {
                    res.add(val*10 + add);
                }
            }
        }

        return res;
    }
}