grid47 Exploring patterns and algorithms
Aug 19, 2024
6 min read
Solution to LeetCode 797: All Paths From Source to Target Problem
You are given a directed acyclic graph (DAG) with n nodes, labeled from 0 to n-1. Find all possible paths from node 0 to node n-1 and return these paths in any order. The graph is represented such that each node has a list of other nodes that can be visited directly from it.
Problem
Approach
Steps
Complexity
Input: The input is a list of lists, where `graph[i]` contains all the nodes you can visit directly from node `i`.
Example: Input: graph = [[1, 2], [3], [3], []]
Constraints:
• 2 <= n <= 15
• 0 <= graph[i][j] < n
• graph[i][j] != i (no self-loops)
• All elements in graph[i] are unique
• The input graph is guaranteed to be a DAG
Output: The output should be a list of all possible paths from node 0 to node n-1. Each path is represented as a list of nodes, starting from node 0 and ending at node n-1.
Example: Output: [[0, 1, 3], [0, 2, 3]]
Constraints:
• The answer will be a list of lists, where each list represents a valid path from node 0 to node n-1.
Goal: The goal is to find all paths starting from node 0 to node n-1, respecting the directed edges in the graph.
Steps:
• Use a breadth-first search (BFS) or depth-first search (DFS) approach to explore the graph starting from node 0.
• For each node, explore all possible nodes it can lead to by following the directed edges.
• Once you reach node n-1, add the path to the result.
• If a node cannot lead to node n-1, backtrack and explore other paths.
Goal: Ensure that the graph follows the specified constraints.
Steps:
• The graph must be a valid directed acyclic graph (DAG).
• Each node has a unique set of nodes it can lead to.
• The graph must contain at least two nodes.
Assumptions:
• The graph is non-cyclic, meaning it does not contain any cycles.
• The graph has at least two nodes (i.e., node 0 and node n-1).
• Input: Input: graph = [[1, 2], [3], [3], []]
• Explanation: In this graph, node 0 can reach nodes 1 and 2, node 1 can reach node 3, and node 2 can also reach node 3. The two possible paths from node 0 to node 3 are [0, 1, 3] and [0, 2, 3].
• Explanation: This graph has multiple paths from node 0 to node 4: [0, 4], [0, 3, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4], and [0, 1, 4].
Approach: A depth-first search (DFS) approach is used to explore all possible paths from node 0 to node n-1. Starting from node 0, recursively visit each neighboring node, adding it to the current path. When node n-1 is reached, add the path to the result.
Observations:
• We need to explore all possible paths in the DAG from node 0 to node n-1.
• DFS or BFS can be used for exploring the graph, as both approaches allow us to find all paths.
Steps:
• Initialize a stack or queue to hold the current node and path.
• Start DFS from node 0 with the path containing only node 0.
• For each node, recursively explore all its neighbors.
• When reaching node n-1, add the current path to the result list.
Empty Inputs:
• If the graph is empty or contains no edges, no paths can be found.
Large Inputs:
• If the graph is very large, ensure the solution can handle it efficiently, particularly considering the constraint of n <= 15.
Special Values:
• If all nodes except for node 0 have no outgoing edges, there will be no paths from node 0 to node n-1.
Constraints:
• The graph must have at least two nodes (node 0 and node n-1).
The function starts by defining the return type as a 2D vector of integers, which will store all the possible paths from the source to the target.
2 : Variable Initialization
vector<vector<int>> ans;
This line initializes a 2D vector 'ans' which will store the valid paths found from source to target.
3 : Queue Initialization
queue<pair<int, vector<int>>> q;
A queue is initialized to facilitate BFS traversal. Each element in the queue is a pair consisting of a node index and a partial path leading up to that node.
4 : Initial Push
q.push({0, {0}});
The starting node (0) is pushed onto the queue, along with an initial path that only contains the source node.
5 : While Loop
while(!q.empty()) {
This while loop continues until the queue is empty, processing each node in the graph in a breadth-first manner.
6 : Extract Node
int x = q.front().first;
The node index 'x' is extracted from the front of the queue.
7 : Extract Path
vector<int> t = q.front().second;
The path leading to node 'x' is extracted from the front of the queue.
8 : Pop Queue
q.pop();
The front element (node and path) is removed from the queue to continue processing the next element.
9 : Check Target
if(x == graph.size() -1) ans.push_back(t);
If the current node 'x' is the target node, the current path 't' is added to the result set 'ans'.
10 : For Loop
for(intk: graph[x]) {
This for loop iterates over all the neighbors 'k' of the current node 'x'.
11 : Push Neighbor
t.push_back(k);
The neighbor node 'k' is added to the current path 't'.
12 : Queue Push
q.push({k, t});
The new pair, consisting of the neighbor 'k' and the updated path 't', is pushed onto the queue.
13 : Pop Last Node
t.pop_back();
After processing the neighbor, the last node is removed from the path 't' to backtrack.
14 : Return Result
return ans;
The function returns the 2D vector 'ans', which contains all the valid paths from the source to the target node.
Best Case: O(n), where n is the number of nodes in the graph.
Average Case: O(2^n), considering that in the worst case, there can be an exponential number of paths.
Worst Case: O(2^n), where n is the number of nodes, since each node could potentially lead to multiple paths.
Description: The time complexity is exponential because we need to explore all possible paths from node 0 to node n-1.
Best Case: O(n), as the space used depends on the recursion depth and the number of paths stored.
Worst Case: O(n), where n is the number of nodes, for storing the current path during DFS.
Description: The space complexity is linear in the worst case, due to recursion and storing the paths.
