Lc 753 Cracking The Safe

Eugene Low Eugene Low Follow Dec 04, 2020 · 6 mins read
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LC 753

753 Cracking the Safe


Euler Path/Circuit

Given a graph, euler path is the path that goes through all edges.

Euler Circuit connects the start and ends of the Euler Path.

This is very similar to topological sort, but it’s harder than topological sort, because the two adjacent result must be connected by an edge.

The graph is suppose to be strongly connected for all the nodes with at least one edge.

For undirected graph:

Euler Path exist if (0/2) # of nodes have odd number of edges.

Euler Circuit exist if all nodes have even number of edges.

For directed graph:

Euler Path exist if all other nodes have same number of i/o edges, expect one node have extra in edges and another have extra out edges.

Euler Circuit exist if all nodes have same number of i/o edges.

Hierholzer’s algorithm

initialise a stack
initialise a deque

    if graph is a euler circuit then push any node
    else (the graph would be euler path) push the startNode (which is the node with odd number of edges or 1 extra in edge)

while stack is not empty:
    top =
    if outEdges of top is zero:
    	deque.push_front( top )
    	for each edge:
        	if edge is not visited:


Given a safe, which can be opened by a n length of password of k characters.

You can insert an arbitary length of a string; in which if the string contains the password, the safe is cracked.

Find the string that can crack all safe with a minimum length.

\[n \in [1,4]\] \[k \in [1,9]\]


input   output
n=1 k=1   0
n=1 k=4   0123
n=2 k=1   00
n=2 k=2   001101
n=3 k=2   00011100101

Thought Process

The first string we can think of which is the worst solution is to concatenate all combination password as one.

n=2 k=2 then 00011011 can be the solution.

Next step is that can see that if we merge two password that have parts that are the same. Ie. 00 and 01 have the same 0 one at the front and another at the back.

In which we can see that 00 and 01 are like the edge between different nodes from front to back.

But what is the node though?

When n is 3

The ...010... password would have the 01 as the front and 10 as the back.

The ...001... password would have the 00 as the front and 01 as the back.

Therefore the fronts and the backs are the nodes!

Indeed if we extend our substring abit longer ie....0011100... We can see that we are actually moving from 00 -> 01 -> 11 -> 11 -> 10 -> 00

We can save space if we keep moving on the graph and not to start from some other place. Indeed if we can find the euler path that will be the optimal solution.

Recall that a graph contains a euler path if it satisfy

Euler Path exist if all other nodes have same number of i/o edges, expect one node have extra in edges and another have extra out edges.

Since all node have k edges, the euler path requirements is satisfied!

So we can just perform the euler path search.


  1. initialise some values:
    1. visited boolean array of False - edge visited
    2. out int array of k - out edges of nodes.
    3. ten_to_n - 10^n
    4. stack - we perform our operation on the item of stack 1. put node 0000(which is same as 0) into stack
    5. path - our result
  2. while stack is not empty:
    1. if top of the stack does not have out edges (out[top] == 0):
      1. pop the top of the stack
      2. add top to path
    2. else: 1. dont pop the top of the stack
      1. for each out edge:
        1. if not edge is not visited: 1. set edge visited to True
          1. reduce out of the node(top) by one (out[top]--)
          2. add the node point by the other side of the edge to stack
          3. break
  3. the reverse of the path is the solution


class Solution:
    def crackSafe(self, n: int, k: int) -> str:
        if n == 1:
            return ''.join(map(str,range(k)))
        visited = [False] * (10**4);
        out = [k] * (10**4);
        ten_to_n = 10**n;
        stack = [0]
        path = []
        while stack:
            # print(stack)
            top = stack[-1];
            if out[top] == 0:
                for edgeMod in range(k):
                    edge = (top * 10) % ten_to_n + edgeMod
                    if not visited[edge]:
                        visited[edge] = True;
                        out[top] -= 1;
        path = list(reversed(path))
        return str(path[1]).zfill(n) + ''.join(map(lambda x: str(x % 10), path[2:]))
Eugene Low
Written by Eugene Low
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