753. Cracking the Safe

Question

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

You can keep inputting the password, the password will automatically be matched against the last n digits entered.

For example, assuming the password is “345”, I can open it when I type “012345”, but I enter a total of 6 digits.

Please return any string of minimum length that is guaranteed to open the box after the entire string is inputted.

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.

Solutions

Method 1: DFS: Hamiltonion path.

Hamilton path question is visiting all of the states without repeating.
What’s the path between the states in this question?
- For each state, its length is n, and next node can re-use previous node’s last n - 1 characters.
- for example, we have 0 and 1 to use.
- first node is 00, second reuses 0 and append 0 or 1.

Analyse this question

If we cannot re-use previous node, we have total length of (k ^ n) * n

Since we can re-use previous node’s n - 1 character, we have the total length of k ^ n + n - 1

  class Solution {
  private int expect; // total length of result.
  private int k;
  private int n;
  public String crackSafe(int n, int k) {
  this.expect = (int)Math.pow((double)k, (double)n) + n - 1;
  this.k = k;
  this.n = n;
  StringBuilder res = new StringBuilder();
  for(int i = 0; i < n; i++) res.append(0);
  Set<String> visited = new HashSet<>();
  visited.add(res.toString());
  dfs(res, visited);
  return res.toString();
  }
  private boolean dfs(StringBuilder sb, Set<String> visited){
  if(sb.length() == expect) return true;
  // sb currently saves the string and we will reuse last n - 1 characters.
  String pre = sb.substring(sb.length() - n + 1);
  for(char c = '0'; c < '0' + k; c++){
      String current = pre + c;
      if(visited.contains(current)) continue;
      visited.add(current);
      sb.append(c);
      if(dfs(sb, visited)) return true;
      sb.deleteCharAt(sb.length() - 1);
      visited.remove(current);
  }
  return false;
  }
  }