leetcode_gym/_problems/allPathsFromSourceToTarget.md

layout, title, last_solved, category, leetcode_url, status

layout	title	last_solved	category	leetcode_url	status
page	All Paths from Source To Target	2020-07-24	graph, bfs, dfs	https://leetcode.com/problems/all-paths-from-source-to-target/	Attempted

Problem

Given a directed, acyclic graph of N nodes.  Find all possible paths from node 0 to node N-1, and return them in any order.

The graph is given as follows:  the nodes are 0, 1, ..., graph.length - 1.  graph[i] is a list of all nodes j for which the edge (i, j) exists.

Example:
Input: [[1,2], [3], [3], []] 
Output: [[0,1,3],[0,2,3]] 
Explanation: The graph looks like this:
0--->1
|    |
v    v
2--->3
There are two paths: 0 -> 1 -> 3 and 0 -> 2 -> 3.

Solution

Timothy's Change DFS

{% highlight python %}

def allPathsSourceTarget(self, graph: List[List[int]]) -> List[List[int]]:

end = len(graph)-1

def dfs(node, path, output):
    
    # base condition
    if node == end:
        output.append(path)
            
    for nxt in graph[node]:
        dfs(nxt, path+[nxt], output)

output = []
dfs( 0, [0], output)
return output

{% endhighlight %}

My Attempt at DFS

{% highlight python %}

def allPathsSourceTarget(self, graph: List[List[int]]) -> List[List[int]]:

output = []

def dfs(node, path, output):    # node is current Node 0,1,2,3
    
    # base condition
    if node==len(graph)-1:
        output.append(path)
    
    else:
        
        # recurse dfs on all children/neighbours
        for neighbour in graph[node]:   
            dfs(neighbour, path + [neighbour], output)

dfs(0, [0], output)
return output

def allPathsSourceTarget(self, graph: List[List[int]]) -> List[List[int]]:

output = []

def dfs(node, path):    # node is current Node 0,1,2,3
    
    #nonlocal output
    
    # base condition
    if node==len(graph)-1:
        output.append(path)
    
    else:
        
        # recurse dfs on all children/neighbours
        for neighbour in graph[node]:   
            dfs(neighbour, path + [neighbour])

dfs(0, [0])
return output

{% endhighlight %}

______________-

BFS SOlution

Anish

{% highlight python %}

def allPathsSourceTarget(self, graph: List[List[int]]) -> List[List[int]]:
    
    q = [[0]]
    
    output = []
    
    end = len(graph)-1
    
    while q:
        
        curr = q.pop(0)
        
        # base condition
        if curr[-1]==end:
            output.append(curr)
        
        else:
            
            for neighbour in graph[curr[-1]]:
                q.append(curr + [neighbour])
        
    return output

{% endhighlight %}