Cracking The Safe Problem

Description

LeetCode Problem 753.

There is a safe protected by a password. The password is a sequence of n digits where each digit can be in the range [0, k - 1].

The safe has a peculiar way of checking the password. When you enter in a sequence, it checks the most recent n digits that were entered each time you type a digit.

• For example, the correct password is “345” and you enter in “012345”:
• After typing 0, the most recent 3 digits is “0”, which is incorrect.
• After typing 1, the most recent 3 digits is “01”, which is incorrect.
• After typing 2, the most recent 3 digits is “012”, which is incorrect.
• After typing 3, the most recent 3 digits is “123”, which is incorrect.
• After typing 4, the most recent 3 digits is “234”, which is incorrect.
• After typing 5, the most recent 3 digits is “345”, which is correct and the safe unlocks.

Return any string of minimum length that will unlock the safe at some point of entering it.

Example 1:

``````1
2
3
Input: n = 1, k = 2
Output: "10"
Explanation: The password is a single digit, so enter each digit. "01" would also unlock the safe.
``````

Example 2:

``````1
2
3
4
5
6
7
8
Input: n = 2, k = 2
Output: "01100"
Explanation: For each possible password:
- "00" is typed in starting from the 4^th digit.
- "01" is typed in starting from the 1^st digit.
- "10" is typed in starting from the 3^rd digit.
- "11" is typed in starting from the 2^nd digit.
Thus "01100" will unlock the safe. "01100", "10011", and "11001" would also unlock the safe.
``````

Constraints:

• 1 <= n <= 4
• 1 <= k <= 10
• 1 <= k^n <= 4096

Sample C++ Code

``````1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
class Solution {
// Brute force idea. try to append every number from 0 to k - 1.
// If we got k^n pwd, then we done.
// Use a map to store which pwd has is visited and pruning.
// If we got same pwd after append a number, this sequence must not be shortest.
bool dfs(const int n, const int k, unordered_map<string, bool>&vis, string& ans) {
if (vis.size() == pow(k, n)) {
return true;
}
for (int i = 0; i < k; ++i) {
ans += '0' + i;
string pwd = ans.substr(ans.length() - n);
if (vis.find(pwd) == vis.end()) {
vis[pwd] = 1;
if (dfs(n, k, vis, ans)) {
return true;
}
vis.erase(pwd);
}
ans.pop_back();
}
return false;
}
public:
string crackSafe(int n, int k) {
string ans = string(n, '0');
unordered_map<string, bool>vis;
vis[ans] = 1;
dfs(n, k, vis, ans);
return ans;
}
};
``````