# 753. Cracking the Safe

## Description

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:

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:

Input: n = 2, k = 2
Output: "01100"
- "00" is typed in starting from the 4th digit.
- "01" is typed in starting from the 1st digit.
- "10" is typed in starting from the 3rd digit.
- "11" is typed in starting from the 2nd digit.
Thus "01100" will unlock the safe. "01100", "10011", and "11001" would also unlock the safe.


Constraints:

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

## Solutions

• class Solution {
private Set<Integer> vis = new HashSet<>();
private StringBuilder ans = new StringBuilder();
private int mod;

public String crackSafe(int n, int k) {
mod = (int) Math.pow(10, n - 1);
dfs(0, k);
ans.append("0".repeat(n - 1));
return ans.toString();
}

private void dfs(int u, int k) {
for (int x = 0; x < k; ++x) {
int e = u * 10 + x;
int v = e % mod;
dfs(v, k);
ans.append(x);
}
}
}
}

• class Solution {
public:
string crackSafe(int n, int k) {
unordered_set<int> vis;
int mod = pow(10, n - 1);
string ans;
function<void(int)> dfs = [&](int u) {
for (int x = 0; x < k; ++x) {
int e = u * 10 + x;
if (!vis.count(e)) {
vis.insert(e);
dfs(e % mod);
ans += (x + '0');
}
}
};
dfs(0);
ans += string(n - 1, '0');
return ans;
}
};

• class Solution:
def crackSafe(self, n: int, k: int) -> str:
def dfs(u):
for x in range(k):
e = u * 10 + x
if e not in vis:
v = e % mod
dfs(v)
ans.append(str(x))

mod = 10 ** (n - 1)
vis = set()
ans = []
dfs(0)
ans.append("0" * (n - 1))
return "".join(ans)


• func crackSafe(n int, k int) string {
mod := int(math.Pow(10, float64(n-1)))
vis := map[int]bool{}
ans := &strings.Builder{}
var dfs func(int)
dfs = func(u int) {
for x := 0; x < k; x++ {
e := u*10 + x
if !vis[e] {
vis[e] = true
v := e % mod
dfs(v)
ans.WriteByte(byte('0' + x))
}
}
}
dfs(0)
ans.WriteString(strings.Repeat("0", n-1))
return ans.String()
}