Leetcode #753: Cracking the Safe
In this guide, we solve Leetcode #753 Cracking the Safe in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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].
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Depth-First Search, Graph, Eulerian Circuit
Intuition
The data forms a graph, so we should explore nodes and edges systematically.
A traversal ensures we visit each node once while maintaining the needed state.
Approach
Build an adjacency list and traverse with BFS or DFS.
Aggregate results as you visit nodes.
Steps:
- Build the graph.
- Traverse with BFS/DFS.
- Accumulate the required output.
Example
Input: n = 1, k = 2
Output: "10"
Explanation: The password is a single digit, so enter each digit. "01" would also unlock the safe.
Python Solution
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)
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.