使用数字 0 到 D 包含所有可能的 N 长度排列的最小数字

给定两个整数N和K ，任务是找到包含长度为N 的所有排列的最小字符串的大小，这些排列可以使用前D 个数字(0, 1, …, D-1) 形成。

例子：

Input: N = 2, D = 2
Output: 01100
Explanation:
Possible permutations of length 2 from digits (0, 1) are {00, 01, 10, 11}.
“01100” is one such string that contains all the permutations as a substring.
Other possible answers are “00110”, “10011”, “11001”

Input: N = 2, D = 4
Output: 03322312113020100
Explaination:
Here all possible permutations of length 2 from digits {0, 1, 2, 3} are
00 10 20 30
01 11 21 31
02 12 22 32
03 13 23 33
“03322312113020100” is a string of minimum length that contains all the above permutations.

方法：
追加 ‘0’ N-1 次并在当前状态的字符串上调用 DFS。将所有 D字符一一附加。每次追加后，检查新字符串是否被访问。如果是这样，通过将其插入 HashSet 并在答案中添加此字符来标记它已访问。在最后的 D 个字符上递归调用 DFS函数。重复这个过程，直到 D 位数字中所有可能的长度为 N 的子串都附加到字符串。打印生成的最终字符串。

下面是上述方法的实现：

Java

// Java Program to find the
// minimum length string
// consisting of all
// permutations of length N
// of D digits
import java.io.*;
import java.util.*;
import java.lang.*;
  
class GeeksforGeeks {
  
    // Initialize hashset to see
    // if all the possible
    // permutations are present
    // in the min length string
    static Set visited;
    // To keep min length string
    static StringBuilder ans;
  
    public static String reqString(int N,
                                   int D)
    {
        // Base case
        if (N == 1 && D == 1)
            return "0";
        visited = new HashSet<>();
        ans = new StringBuilder();
  
        StringBuilder sb = new StringBuilder();
        // Append '0' n-1 times
        for (int i = 0; i < N - 1; ++i) {
            sb.append("0");
        }
        String start = sb.toString();
        // Call the DFS Function
        dfs(start, D);
        ans.append(start);
  
        return new String(ans);
    }
    // Generate the required string
    public static void dfs(String curr, int D)
    {
        // Iterate over all the possible
        // character
        for (int x = 0; x < D; ++x) {
            // Append to make a new string
            String neighbour = curr + x;
            // If the new string is not
            // visited
            if (!visited.contains(neighbour)) {
                // Add in hashset
                visited.add(neighbour);
                // Call the dfs function on
                // the last d characters
                dfs(neighbour.substring(1), D);
  
                ans.append(x);
            }
        }
    }
    // Driver Code
    public static void main(String args[])
    {
  
        int N = 2;
        int D = 2;
        System.out.println(reqString(N, D));
    }
}

输出：

01100

时间复杂度： O(N * D ^N )
辅助空间： O(N * D ^N )

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