753. Cracking the Safe

HardDepth-First SearchGraphEulerian Circuit

Problem Description

In this problem, we're tasked with finding a string that will eventually unlock a safe that is protected by a password. The password itself is a sequence of n digits, and each digit ranges from 0 to k - 1. The safe uses a sliding window to check the correctness of the entered password, comparing the most recent n digits entered against the actual password.

For instance, if the password of the safe is "345" (n=3), and you enter the sequence "012345", the safe checks "0", "01", "012", "123", "234", and finally "345". Only the last sequence "345" matches the password, so the safe unlocks.

The goal is to generate the shortest possible string such that, as we type in this string, the safe will eventually identify the correct password within the most recent n digits and unlock.

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

Yes: The problem involves finding a sequence that includes all possible combinations of a given length n of digits k, which can be visualized as a graph where each node represents a state (specific combination of digits), and edges represent transitions between states.

Is it a tree?

No: The graph is not hierarchical and can contain cycles, meaning each combination can potentially be reached from multiple other combinations.

Is the problem related to directed acyclic graphs (DAGs)?

No: Although the graph can be directed (going from one combination to another), it is not acyclic because it's possible to return to a previous combination after visiting all combinations (due to the nature of the problem, where every possible combination needs to be used exactly once).

Is the problem related to shortest paths?

No: The target is not to find the shortest path but to ensure all possible paths are covered.

Does the problem involve connectivity?

Yes: We need to ensure that the entire set of combinations (graph) is covered without repetitions.

Does the problem have small constraints?

Yes: Because the total number of combinations is ( k^n ), where both k (the number of digits) and n (length of each combination) tend to be small enough in practical problems to allow complete exploration.

Conclusion: The flowchart suggests using DFS/backtracking for this problem, as it involves exploring all paths in a graph with small constraints involving connectivity. This approach will help to systematically traverse and record each unique path (combination) once, ensuring that all combinations (graph nodes) are visited.

Intuition

The solution to this problem uses a depth-first search (DFS) algorithm to build a sequence that contains every possible combination of n digits with digits from 0 to k - 1. This is akin to finding an Eulerian path or circuit in a graph, where each node represents a (n-1)-digit sequence, and each directed edge represents a possible next digit that can be added to transform one (n-1)-digit sequence into a n-digit sequence.

The intuition is to visit each possible combination (every 'node') exactly once. Starting with the sequence "0" repeated n-1 times, the algorithm adds one digit at a time by traversing unvisited edges (using the remaining k possible digits), and then moves onto the next sequence by removing the first digit and appending a new one, much like in a sliding window approach.

A hash set (vis) is used to keep track of visited combinations, ensuring that each possible n-digit combination is entered exactly once. The mod operation (e % mod) in the DFS function is used to obtain the next (n-1)-digit sequence (the 'node') after adding a new digit. The DFS continues until all possible sequences are visited.

At the end, the function returns the constructed string which will include the necessary sequence to unlock the safe at some point when typed in.

Learn more about Depth-First Search and Graph patterns.

Solution Approach

The solution uses a Depth-First Search (DFS) algorithm to explore the sequences of digits that will unlock the safe. Here's an explanation of the implementation:

Depth-First Search (DFS) Function: The core of the solution is the dfs function, which attempts to visit all n-digit sequences exactly once. Each call to dfs handles a particular (n-1)-digit sequence (u), trying to append each of the k possible next digits (x) to form a n-digit sequence.
Handling Combinations: Each time we consider adding a digit x to sequence u, we form a new sequence (e) by doing u * 10 + x. This operation essentially shifts the digits left and appends the new digit to the right.
Visited Sequences Tracking: A set named vis is used to keep track of visited n-digit sequences. If a sequence e has not been visited yet (e not in vis), it's added to the vis set to mark it as visited.
Sliding to Next Sequence: After marking e as visited, we calculate the next (n-1)-digit sequence v to continue the DFS. This is done via e % mod, where mod = 10 ** (n - 1). The modulo operation essentially removes the leftmost digit from e (because we've already handled this n-digit sequence).
Building the Result: As we explore each digit x that can be added to u, we append x to ans, which is a list that will eventually contain all digits of the minimum length string necessary to unlock the safe.
Initialization and Starting the DFS: We initialize ans and vis, and then we start the DFS from the initial sequence "0" repeated n-1 times (dfs(0)). This assumes that the preceding sequence of digits ends with zeros, which is a neutral assumption since we seek any valid sequence to unlock the safe.
Finalizing the Result: Once the DFS is complete, we append "0" repeated n-1 times to ans to represent the initial state where the safe started checking the digits. We leave the DFS function with DFS(0), which inputs the first valid n-1 digit combination.
Concatenating the Digits: Finally, we join all the digits in ans together into a string ("".join(ans)) and return this string as the solution.

This approach guarantees that we generate a string with all possible n-digit combinations that can occur as the safe checks the last n digits entered. It ensures that somewhere in this string is the correct sequence to unlock the safe, effectively solving the problem.

Ready to land your dream job?

Unlock your dream job with a 2-minute evaluator for a personalized learning plan!

Start Evaluator

Example Walkthrough

Let's illustrate the solution approach using an example. Suppose the password of the safe is a sequence of n = 2 digits, and each digit may range from 0 to k - 1 where k = 2. This means each digit in the password can either be 0 or 1.

Now let's walk through how the DFS algorithm would find the shortest possible string that includes the password:

Initialization: We initialize ans as an empty list to store our result, and vis as an empty set to track visited sequences. We're aiming to start DFS with an initial sequence of n-1 digit, which is "0".
Starting with "0": We start with the sequence "0", and we will try to append 0 and 1 to it in the DFS.
DFS from "0": During the DFS, we try appending each digit.
- First, we append 0 to "0" making it "00".
- Since "00" hasn't been visited, we mark it as visited and then the DFS considers "0" and tries to append 1.
- Next, we append 1 to "0" which gives us "01".
- Again, "01" hasn't been visited, so we mark it and consider the next sequence which only contains "1".
DFS from "1": Now our sequence is "1", and we repeat the process.
- We try appending 0 to "1", yielding "10".
- "10" is not visited, hence we mark it as visited and add it to ans.
- Now we try appending 1 to "1", resulting in "11".
- "11" is not in vis. We mark it as visited and it's traced in ans.
Tracking visited combinations: At this point, all possible n-digit combinations have been visited: "00", "01", "10", "11".
Building the Result: Our ans list consists of the digits we added to sequences in the DFS order: [0, 1, 0, 1].
Final Sequence: To complete the string, we need to consider an initial state represented by n-1 zeros. Since n=2, we add one zero to the beginning: 0 + ans = [0, 0, 1, 0, 1].
Concatenating into a String: We join all the digits in ans to form the string "00101".

This string "00101" is the shortest sequence that ensures the safe will unlock, as it includes all possible combinations of the n digits "00", "01", "10", "11" within its consecutive characters. If the actual password were "00", "01", "10", or "11", it would be found within this string as the safe's sliding window checks the most recent n digits entered.

Solution Implementation

1class Solution:
2    def crackSafe(self, n: int, k: int) -> str:
3        # Depth-first search function to traverse through the nodes/vertices
4        def dfs(current_vertex):
5            for x in range(k):
6                edge = current_vertex * 10 + x  # Forming a new edge
7                if edge not in visited:  # If the edge is not visited
8                    visited.add(edge)  # Mark the edge as visited
9                    next_vertex = edge % modulus  # Calculate the next vertex
10                    dfs(next_vertex)  # Recursive call to visit the next vertex
11                    answer.append(str(x))  # Append the current character to the answer
12
13        # Calculate the modulus for finding vertices
14        modulus = 10 ** (n - 1)
15        # Initialize a set to keep track of visited edges
16        visited = set()
17        # List to store the characters of the answer
18        answer = []
19        dfs(0)  # Start DFS from the vertex 0
20        # Append the initial combination to the answer (n-1 zeros)
21        answer.append("0" * (n - 1))
22        # Join all characters to form the final answer string and return
23        return "".join(answer)
24

1class Solution {
2    // We use a HashSet to keep track of visited nodes during DFS
3    private Set<Integer> visited = new HashSet<>();
4  
5    // StringBuilder to store the answer
6    private StringBuilder answer = new StringBuilder();
7  
8    // The modulo for trimming the prefix when moving to the next node
9    private int modulus;
10
11    /**
12     * Starts the process to find the sequence that cracks the safe.
13     *
14     * @param n the number of digits in the safe's combination
15     * @param k the range of digits from 0 to k-1 that can be used in the combination
16     * @return a string representing the minimum length sequence that cracks the safe
17     */
18    public String crackSafe(int n, int k) {
19        // Calculate the modulus to use for creating n-1 length prefixes
20        modulus = (int) Math.pow(10, n - 1);
21      
22        // Start DFS from node 0
23        dfs(0, k);
24      
25        // Append the initial prefix to complete the sequence
26        for (int i = 0; i < n - 1; i++) {
27            answer.append("0");
28        }
29        return answer.toString();
30    }
31
32    /**
33     * Performs DFS to find the Eulerian path/circuit in the De Bruijn graph.
34     *
35     * @param node the current node in the graph
36     * @param k the range of digits from 0 to k-1
37     */
38    private void dfs(int node, int k) {
39        // Try all possible next digits to create an edge from the current node
40        for (int x = 0; x < k; ++x) {
41            // Create the new edge by appending digit x to the current node
42            int edge = node * 10 + x;
43            // If the edge has not been visited, add it to the visited set
44            if (visited.add(edge)) {
45                // Calculate the next node by removing the oldest digit (modulus)
46                int nextNode = edge % modulus;
47                // Perform DFS on the next node
48                dfs(nextNode, k);
49                // Append the current digit to the answer
50                answer.append(x);
51            }
52        }
53    }
54}
55

1#include <functional>
2#include <unordered_set>
3#include <string>
4#include <cmath>
5
6class Solution {
7public:
8    // Function to generate and return the De Bruijn sequence.
9    // 'n' represents the length of the subsequences and 
10    // 'k' represents the range of digits (0 to k-1).
11    string crackSafe(int n, int k) {
12        // Create a set to keep track of visited combinations.
13        std::unordered_set<int> visited;
14        // Compute 10^(n-1), used later to find the next state.
15        int mod = std::pow(10, n - 1);
16        // Initialize the answer string.
17        string result;
18
19        // Depth-first search function to explore all possible combinations.
20        // 'u' represents the current combination being explored as a prefix.
21        std::function<void(int)> dfs = [&](int u) {
22            // Try to append each possible digit from 0 to k-1.
23            for (int x = 0; x < k; ++x) {
24                // Create the new mixed radix combination by appending the digit x.
25                int combo = u * 10 + x;
26                // If the combination is not yet visited, continue the exploration.
27                if (visited.count(combo) == 0) {
28                    // Mark the new combination as visited.
29                    visited.insert(combo);
30                    // Recursively explore the next state by taking the last (n-1) digits.
31                    dfs(combo % mod);
32                    // Append the current digit to the result.
33                    result.push_back(x + '0');
34                }
35            }
36        };
37
38        // Start the DFS from the combination of n zeroes.
39        dfs(0);
40        // To close the De Bruijn sequence, append n-1 zeroes at the end.
41        result += std::string(n - 1, '0');
42        return result;
43    }
44};
45

1// Importing necessary utilities from external libraries
2// is not needed in TypeScript for the described functionality.
3
4// Define the type for our Depth-first search (DFS) function
5type DFSFunction = (u: number) => void;
6
7// Function to generate and return the De Bruijn sequence.
8// Parameter 'n' represents the length of the subsequences,
9// and parameter 'k' represents the range of digits (0 to k-1).
10function crackSafe(n: number, k: number): string {
11    // Set to keep track of visited combinations.
12    let visited: Set<number> = new Set();
13
14    // Compute 10^(n-1), used later to find the next state.
15    let mod: number = Math.pow(10, n - 1);
16
17    // Initialize the answer string.
18    let result: string = "";
19
20    // Depth-first search function to explore all possible combinations.
21    let dfs: DFSFunction = (u: number) => {
22        // Try to append each possible digit from 0 to k-1.
23        for (let x = 0; x < k; ++x) {
24            // Create the new mixed radix combination by appending the digit x.
25            let combo: number = u * 10 + x;
26            // If the combination is not yet visited, continue the exploration.
27            if (!visited.has(combo)) {
28                // Mark the new combination as visited.
29                visited.add(combo);
30                // Recursively explore the next state by taking the last (n-1) digits.
31                dfs(combo % mod);
32                // Append the current digit to the result.
33                result += x.toString();
34            }
35        }
36    }
37
38    // Start the DFS from the combination of n zeroes.
39    dfs(0);
40
41    // To close the De Bruijn sequence, append n-1 zeroes at the end.
42    result += '0'.repeat(n - 1);
43
44    return result;
45}
46
47// The crackSafe function can now be used globally in any TypeScript file
48

Time and Space Complexity

The given Python code aims to generate the shortest string that contains all possible combinations of a given length n using digits from 0 to k-1, which is essentially the De Bruijn sequence problem. In essence, the algorithm performs a Depth-First Search (DFS) to construct the Eulerian circuit/path in a De Bruijn graph.

Time Complexity

The time complexity can be analyzed based on the total number of nodes and edges visited in the DFS. Each node represents a unique combination of n-1 digits, resulting in k^(n-1) possible nodes. The DFS visits each edge exactly once. Since the graph is a directed graph with k possible outgoing edges from each node, there will be a total of k^n edges.

Therefore, the time complexity of the DFS is O(k^n), as this is the number of edges, and each edge is visited once.

Space Complexity

The space complexity is primarily determined by the storage of the visited edges, which is managed by the vis set. Since each unique combination of n digits is stored as an edge, up to k^n edges can be in the set, leading to a space complexity of O(k^n).

The function also uses a recursive call stack for DFS, which in the worst case could grow up to the number of edges, i.e., O(k^n) in space.

The ans list stores all the visited edges' last digits plus the padding at the end, accounting for at most k^n + (n-1) elements. However, since k^n dominates n-1 as k^n can grow much larger with increasing n and k, the contribution of n-1 can be considered negligible for large enough k and n.

Overall, the space complexity is O(k^n), dominated by the storage requirements of the vis set and the recursion call stack.

Learn more about how to find time and space complexity quickly using problem constraints.

Discover Your Strengths and Weaknesses: Take Our 2-Minute Quiz to Tailor Your Study Plan: