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Cracking the Safe

Difficulty: Hard

Problem 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 checks the most recent n digits entered each time a digit is typed. Return any string of minimum length that will unlock the safe at some point of entering it.

Key Insights

  • The password consists of n digits, and the range of digits is from 0 to k-1.
  • Each time a digit is entered, the last n digits are checked against the correct password.
  • To ensure all possible combinations of passwords are checked, we can generate a sequence that contains all n-length combinations from the available digits.
  • The problem can be approached using an Eulerian path in a graph, where each node represents a sequence of n-1 digits, and edges represent the next digit added.

Space and Time Complexity

Time Complexity: O(k^n)
Space Complexity: O(k^n)

Solution

To solve this problem, we can use a depth-first search (DFS) approach to generate a string that contains all possible sequences of n digits. We will treat the sequences as nodes in a directed graph where each edge connects to a new digit. The approach ensures that we traverse each combination efficiently, using each sequence of n-1 digits to build the next digit.

  1. Initialize a set to keep track of visited combinations.
  2. Use a DFS to construct the password by appending digits and checking if the last n digits form a valid sequence.
  3. Backtrack if necessary and continue exploring other paths.
  4. The resulting string will be the shortest possible sequence that unlocks the safe.

Code Solutions

def crackSafe(n: int, k: int) -> str:
    # Create the initial password with n-1 zeros
    password = '0' * n
    visited = set()
    
    def dfs(current):
        for x in range(k):
            # Create the next combination
            next_combination = current + str(x)
            # If we haven't visited this combination yet
            if next_combination not in visited:
                visited.add(next_combination)
                # Continue with DFS
                dfs(next_combination[1:])
                # Append the digit to the password
                nonlocal password
                password += str(x)

    visited.add(password)
    dfs(password)
    return password

# Example usage
print(crackSafe(2, 2))  # Output: "01100" or similar
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