753. Cracking the Safe

标签：better   between   cte   lse   pre   git   its   nod   oss   

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:

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

Approach #1: DFS. [Java]

class Solution {
    public String crackSafe(int n, int k) {
        String strPwd = String.join("", Collections.nCopies(n, "0"));
        StringBuilder sbPwd = new StringBuilder(strPwd);
        int total = (int)Math.pow(k, n);
        Set<String> seen = new HashSet<>();
        seen.add(strPwd);
        
        crackSafeAfter(sbPwd, total, seen, n, k);

        return sbPwd.toString();
    }
    
    private boolean crackSafeAfter(StringBuilder pwd, int total, Set<String> seen, int n, int k) {
        if (seen.size() == total) return true;
        
        String lastDigits = pwd.substring(pwd.length()-n+1);
        for (char ch = ‘0‘; ch < ‘0‘ + k; ch++) {
            String newComb = lastDigits + ch;
            if (!seen.contains(newComb)) {
                seen.add(newComb);
                pwd.append(ch);
                if (crackSafeAfter(pwd, total, seen, n, k)) return true;
                seen.remove(newComb);
                pwd.deleteCharAt(pwd.length() - 1);
            }
        }
        
        return false;
    }
}

Analysis:

In order to guarantee to open the box at last, the input password ought to contain all length-n combinations on digits [0...k-1] - there should be k^n combinations in total.

To make the input password as short as possible, we‘d better make each possible length-n combination on digits [0...k-1] occurs exactly once as a substring of the password. The existence of such a password is proved by DeBruijin sequence:

A De Bruijn sequence of order n on a size-k alphabet A is a cyclic sequence in which every possible length-n string on A occurs exactly once as a substring. It has length k^n, which is also the number of distinct substrings of length n on a size-k alphabet; De Bruijn sequences are therefore optimally short.

We reuse last n-1 digits of the input-so-far password as below:

e.g. n = 2, k = 2

all 2-length combinations on [0, 1]:

00 (‘00‘110)

  01 (0‘01‘10)

    11 (00‘11‘0)

　  10 (001‘10‘)

The password is 00110

We can utilize DFS to find the password:

goal: to find the shortest input password such that each possible n-length combination of digits [0..k-1] occurs exactly once as a substring.

node: current input password

edge: if the last n - 1 digits of node1 can be transformed to node2 by appending a digit from 0..k-1, there will be an edge between node1 and node2

start node: n repeated 0‘s
end node: all n-length combinations among digits 0..k-1 are visited

visitedComb: all combinations that have been visited

Reference:

https://leetcode.com/problems/cracking-the-safe/discuss/153039/DFS-with-Explanations

753. Cracking the Safe

标签：better   between   cte   lse   pre   git   its   nod   oss   

原文地址：https://www.cnblogs.com/ruruozhenhao/p/10878535.html