753.Cracking-the-Safe

753. Cracking the Safe

题目地址

https://leetcode.com/problems/cracking-the-safe/

题目描述

There is a box protected by a password. The password is a sequence of n digits where each digit can be one of the first k digits 0, 1, ..., k-1.

While entering a password, the last n digits entered will automatically be matched against the correct password.

For example, assuming the correct password is "345", if you type "012345", the box will open because the correct password matches the suffix of the entered password.

Return any password of minimum length that is guaranteed to open the box at some point of entering it.

Example 1:
Input: n = 1, k = 2
Output: "01"
Note: "10" will be accepted too.

Example 2:
Input: n = 2, k = 2
Output: "00110"
Note: "01100", "10011", "11001" will be accepted too.

Note:
n will be in the range [1, 4].
k will be in the range [1, 10].
k^n will be at most 4096.

代码

Approach #1 Hierholzer's Algorithm

Time: O(N K^N) && Space: O(N K^N)

class Solution {
  Set<String> seen;
  StringBuilder ans;
  public String crackSafe(int n, int k) {
        if (n == 1 && k == 1)        return "0";
    seen = new HashSet();
    ans = new StringBuilder();

    StringBuilder sb = new StringBuilder();
    for (int i = 0; i < n-1; i++)
      sb.append("0");
    String start = sb.toString();

    dfs(start, k);
    ans.append(start);  // post-order
    return new String(ans);
  }

  private void dfs(String node, int k) {
    for (int x = 0; x < k; x++) {
      String nei = node + x;
      if (!seen.contains(nei)) {
        seen.add(nei);
        dfs(nei.substring(1), k);
        ans.append(x);
      }
    }
  }
}

Approach #2 Inverse Burrows-Wheeler Transform Confusion

Time & Space O(k^n)

class Solution {
  public String crackSafe(int n, int k) {
    int M = (int)Math.pow(k, n-1)
    int[] P = new int[M * k];
    for (int i = 0; i < k; i++)
      for (int q = 0; q < M; q++) 
        P[i*M + q] = q * k + i;

    StringBuilder ans = new StringBuilder();
    for (int i = 0; i < M * k; i++) {
      int j = i;
      while (P[j] >= 0) {
        ans.append(String.valueOf(j / M));
        int v = P[j];
        P[j] = -1;
        j = v;
      }
    }

    for (int i = 0; i < n-1; i++)
      ans.append("0");

    return new String(ans);
  }
}