#753

Hard

auto_awesomeGraph traversal with depth-first search

LeetCode Problem Workspace

Cracking the Safe

The Cracking the Safe problem involves finding the shortest password sequence to unlock a safe using graph traversal and Eulerian circuits.

depth first search graph eulerian circuit

Problem Statement

A safe is protected by a password consisting of n digits, where each digit is in the range [0, k-1]. The safe checks the most recent n digits entered each time a new digit is typed in, and you must find the shortest sequence of digits that will eventually unlock the safe.

Given the constraints of the problem, you need to return any valid string of minimum length that unlocks the safe. The challenge is to efficiently traverse through all possible combinations using graph traversal techniques, such as depth-first search and Eulerian circuits, to guarantee the safe will be unlocked.

Examples

Example 1

Input: n = 1, k = 2

Output: "10"

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"

For each possible password:

"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. "10011", and "11001" would also unlock the safe.

Constraints

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

Solution Approach

Graph Traversal with Depth-First Search

This problem can be solved by treating the problem as a graph traversal task, where each state of the password is a node, and the edges represent the transition to the next valid digit. Depth-First Search (DFS) is useful for exploring all possible paths and ensuring that every edge (combination) is visited exactly once, leading to an Eulerian circuit.

Eulerian Circuit Concept

The key idea is that we need to traverse through each possible password sequence exactly once, which aligns with the concept of an Eulerian path. An Eulerian path in a graph visits every edge exactly once and can be applied to solve the problem by treating each possible digit combination as an edge in a directed graph.

Backtracking and Optimization

Backtracking techniques can be applied in DFS to optimize the traversal. Starting from the initial password state, the algorithm will explore all possible transitions and backtrack when necessary, ensuring that the shortest sequence is found while covering all possible combinations of digits.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time and space complexity depend on the specific approach used to solve the problem, but in general, DFS traversal will run in O(kn) time and require O(kn) space, where k is the number of possible digits and n is the length of the password sequence. This is because we need to explore all possible states and keep track of visited paths to avoid revisiting edges.

What Interviewers Usually Probe

The candidate's understanding of Eulerian circuits and graph traversal will be essential for solving this problem.
Look for familiarity with depth-first search and backtracking techniques as key concepts for solving this efficiently.
The ability to break down the problem into smaller graph-related subproblems and use traversal optimizations will be a critical signal of readiness.

Common Pitfalls or Variants

Common pitfalls

Failing to recognize the problem as a graph traversal one, leading to inefficient brute force solutions.
Not considering the constraints properly, which may result in unnecessary revisits to password combinations.
Overlooking the Eulerian circuit approach, which could lead to overly complex or incorrect solutions.

Follow-up variants

Allowing a different number of digits, which could alter the traversal space.
Modifying the range of possible digits (k), potentially requiring a new approach to handle larger spaces.
Changing the sequence length (n), making the problem easier or harder depending on the value of n.

FAQ

How can Depth-First Search help in solving Cracking the Safe?

DFS is critical for exploring all possible paths in Cracking the Safe, ensuring that all combinations are visited exactly once and optimizing the password sequence.

What is an Eulerian circuit, and why is it important for this problem?

An Eulerian circuit is a path that visits every edge in a graph exactly once, which is the key to solving Cracking the Safe, as it ensures every password combination is covered.

What are the constraints for the Cracking the Safe problem?

The constraints are 1 <= n <= 4, 1 <= k <= 10, and 1 <= kn <= 4096, which limit the possible password lengths and digit ranges.

What should I consider when choosing a graph traversal approach for this problem?

Choosing DFS and optimizing with backtracking or Eulerian circuits is important to efficiently traverse the password space without unnecessary re-exploration.

How does GhostInterview assist with solving Cracking the Safe?

GhostInterview provides interactive, graph-focused problem-solving steps that help candidates understand Eulerian circuits, DFS, and optimal traversal methods.

terminal

Solution

Solution 1: Eulerian Circuit

We can construct a directed graph based on the description in the problem: each point is considered as a length $n-1$ $k$-string, and each edge carries a character from $0$ to $k-1$. If there is a directed edge $e$ from point $u$ to point $v$, and the character carried by $e$ is $c$, then the last $k-1$ characters of $u+c$ form the string $v$. At this point, the edge $u+c$ represents a password of length $n$.

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:
                    vis.add(e)
                    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)

Continue Practicing

#332 reconstruct itinerary #743 network delay time #765 couples holding hands