
Problem Statement
In this scenario, we are working with a directed graph characterized by 'n' nodes, where each node has a unique identifier ranging from 0 to n - 1
. The structure of this graph is represented by a 0-indexed 2D integer array named graph
. In this array, graph[i]
holds a list of nodes which direct edges emanate from node i
to each node listed within graph[i]
.
Main concepts:
- Terminal Node: A node with no outbound edges leading to any other node.
- Safe Node: A node is dubbed "safe" if all paths originating from it invariably lead to a terminal node or another safe node.
The problem's goal is to identify and return all the nodes that are considered safe. The output should be an array of these safe nodes, organized in ascending order. This listing helps in verifying circuit designs or network routing configurations where certain safety conditions (deadlock or cycle-free paths) need assurance.
Examples
Example 1
Input:
graph = [[1,2],[2,3],[5],[0],[5],[],[]]
Output:
[2,4,5,6]
Explanation:
The given graph is shown above. Nodes 5 and 6 are terminal nodes as there are no outgoing edges from either of them. Every path starting at nodes 2, 4, 5, and 6 all lead to either node 5 or 6.
Example 2
Input:
graph = [[1,2,3,4],[1,2],[3,4],[0,4],[]]
Output:
[4]
Explanation:
Only node 4 is a terminal node, and every path starting at node 4 leads to node 4.
Constraints
n == graph.length
1 <= n <= 104
0 <= graph[i].length <= n
0 <= graph[i][j] <= n - 1
graph[i]
is sorted in a strictly increasing order.- The graph may contain self-loops.
- The number of edges in the graph will be in the range
[1, 4 * 104]
.
Approach and Intuition
Let's break down the process and underlying intuition through the examples provided:
Understanding Safe and Terminal Nodes:
- A terminal node is straightforward; it's simply any node that doesn't point to another node.
- A safe node, on the other hand, can only be considered safe if every single path that starts from it leads to another safe node or a terminal node.
General Steps to Solve the Problem:
- Identify terminal nodes since these will be part of the output as they inherently meet the safety criteria by not having outbound edges.
- Use depth-first search (DFS), or a similar graph traversal methodology, to track all the nodes from which terminal nodes are reachable.
- During traversal, carefully mark nodes that potentially lead into cycles or nodes that do not eventually lead to terminal nodes. This can leverage marking or coloring nodes during DFS to represent different states (unvisited, visiting, safe).
- Revisit each node, leveraging the markings to confirm if it either stays safe or leads to a terminal node through all possible paths.
By following these steps, we assess each node's connectivity and their eventual paths, thereby filtering out the nodes that inherently satisfy the safety conditions set by the problem statement.
Through examples:
- In Example 1, nodes like 5 and 6 are terminal. Node 2 eventually leads all paths to nodes 5 or 6, hence making it safe. Similar logic confirms the safety of the other nodes in the output.
- In Example 2, the only terminal node is 4. It is also the only safe node since all paths from node 4 lead to itself, meeting the terminal node condition directly.
The intuitive way of solving this problem primarily revolves around efficient graph traversal and a good mechanism for marking nodes based on their connectivity and the destination of their paths. This ensures that all potential pathways are accounted for when establishing the safety of each node.
Solutions
- C++
- Java
- Python
class Solution {
public:
bool cycleDetected(int curr, vector<vector<int>>& graph, vector<bool>& visited,
vector<bool>& recursionStack) {
if (recursionStack[curr]) {
return true;
}
if (visited[curr]) {
return false;
}
visited[curr] = true;
recursionStack[curr] = true;
for (int next : graph[curr]) {
if (cycleDetected(next, graph, visited, recursionStack)) {
return true;
}
}
recursionStack[curr] = false;
return false;
}
vector<int> findSafeNodes(vector<vector<int>>& graph) {
int totalNodes = graph.size();
vector<bool> visited(totalNodes), recursionStack(totalNodes);
for (int i = 0; i < totalNodes; i++) {
cycleDetected(i, graph, visited, recursionStack);
}
vector<int> safeNodesList;
for (int i = 0; i < totalNodes; i++) {
if (!recursionStack[i]) {
safeNodesList.push_back(i);
}
}
return safeNodesList;
}
};
In the provided C++ solution for identifying eventual safe states in a directed graph, the code follows a two-step approach:
Detect cycles within the graph using a modified depth-first search (DFS):
- Implement a
cycleDetected
function that checks if any cycles are present starting from a specific node. It utilizes two vectors to track visited nodes and the call stack (recursionStack) to identify back edges, which signify a cycle. - The function returns
true
when a cycle is detected andfalse
otherwise.
- Implement a
Identify and list all safe nodes:
- A node is defined as "safe" if no cycles can be reached starting from this node.
- After cycling through all nodes with the
cycleDetected
function, inspect each node. Nodes that don't have any active recursionStack marks (indicating that the node does not contribute to a cycle) are added to thesafeNodesList
.
The logic for cycle detection within the graph is critical in filtering out unsafe states, ensuring only truly safe states are added to the resulting list. This approach ensures computational efficiency and correctness in identifying safe states.
class Solution {
public boolean cycleCheck(
int currentNode,
int[][] adjacencyList,
boolean[] visited,
boolean[] recursionStack
) {
if (recursionStack[currentNode]) {
return true; // Cycle found.
}
if (visited[currentNode]) {
return false; // Already processed node.
}
visited[currentNode] = true;
recursionStack[currentNode] = true;
for (int adjacentNode : adjacencyList[currentNode]) {
if (cycleCheck(adjacentNode, adjacencyList, visited, recursionStack)) {
return true;
}
}
recursionStack[currentNode] = false;
return false;
}
public List<Integer> findSafeNodes(int[][] matrix) {
int numVertices = matrix.length;
boolean[] visited = new boolean[numVertices];
boolean[] recursionStack = new boolean[numVertices];
for (int vertex = 0; vertex < numVertices; vertex++) {
cycleCheck(vertex, matrix, visited, recursionStack);
}
List<Integer> safeNodesList = new ArrayList<>();
for (int vertex = 0; vertex < numVertices; vertex++) {
if (!recursionStack[vertex]) {
safeNodesList.add(vertex);
}
}
return safeNodesList;
}
}
This Java solution is designed to determine eventual safe states in a directed graph represented by an adjacency matrix. The central idea hinges on detecting cycles in the graph, as nodes that are part of cycles are not considered safe.
Key Procedure:
Helper Method (
cycleCheck
)- This method uses Depth First Search (DFS) to explore nodes and check for cycles.
- Maintains two boolean arrays,
visited
andrecursionStack
.visited
tracks whether a node has been checked, andrecursionStack
keeps track of nodes currently in the recursion stack (DFS path). - Returns
true
if a cycle is detected; otherwise, it returnsfalse
.
Main Method (
findSafeNodes
)- Initializes arrays to keep track of visited nodes and nodes in the recursion stack.
- Iterates over each vertex and runs the
cycleCheck
for each one. - Constructs a list of safe nodes, i.e., nodes that are not part of any cycle. Nodes not in the recursion stack after checking are considered safe.
- At the conclusion of this approach:
- The code systematically identifies nodes without any outgoing cycles as safe.
- These safe nodes are accumulated in a
List<Integer>
and returned.
This solution adeptly combines graph traversal techniques with cycle detection to flag unsafe nodes efficiently, aligning well with explorations of state safety in graph structures.
class Solution:
def depthFirstSearch(self, current, connections, visited, recursionStack):
if recursionStack[current]:
return True
if visited[current]:
return False
visited[current] = True
recursionStack[current] = True
for next_node in connections[current]:
if self.depthFirstSearch(next_node, connections, visited, recursionStack):
return True
recursionStack[current] = False
return False
def findSafeNodes(self, graph: List[List[int]]) -> List[int]:
total_nodes = len(graph)
visited = [False] * total_nodes
recursionStack = [False] * total_nodes
for index in range(total_nodes):
self.depthFirstSearch(index, graph, visited, recursionStack)
safeNodesList = []
for index in range(total_nodes):
if not recursionStack[index]:
safeNodesList.append(index)
return safeNodesList
The solution revolves around determining safe nodes in a directed graph where a node is termed 'safe' if it does not partake in any cycles. It employs the depth-first search (DFS) method to detect cycles efficiently. Here’s a step-by-step breakdown of the given code logic written in Python:
Define a
depthFirstSearch
method:- Checks for cycle presence when traversing through each node's connections.
- Uses a recursion stack to manage the recursive depth of nodes being visited.
- Marks nodes as visited to prevent redundant checks.
- If a cycle is detected during traversal, it returns
True
, signaling an unsafe node.
Define a
findSafeNodes
method:- Initialize arrays
visited
andrecursionStack
to keep track of visited nodes and the nodes currently in the recursion stack, respectively. - Iterate over each node and apply the
depthFirstSearch
method to identify whether any node is part of a cycle using the stack-based cycle detection. - After validating all nodes, iterate through the
recursionStack
to collect nodes that are not part of any recursion (hence not in a cycle) and append them tosafeNodesList
.
- Initialize arrays
The returned safeNodesList
contains indices of all the nodes which can be classified as safe, meaning there's no path originating from these nodes that leads to a cycle.
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