Monday, April 23, 2018

LeetCode 797 - All Paths From Source to Target


https://leetcode.com/problems/all-paths-from-source-to-target/description/
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.
Note:
  • The number of nodes in the graph will be in the range [2, 15].
  • You can print different paths in any order, but you should keep the order of nodes inside one path.
Since the graph is a directed, acyclic graph, any path from A to B is going to be composed of A plus a path from any neighbor of A to B. We can use a recursion to return the answer.
Algorithm
Let N be the number of nodes in the graph. If we are at node N-1, the answer is just the path {N-1}. Otherwise, if we are at node node, the answer is {node} + {path from nei to N-1} for each neighbor nei of node. This is a natural setting to use a recursion to form the answer.
  • Time Complexity: O(2^N N^2). We can have exponentially many paths, and for each such path, our prepending operation path.add(0, node) will be O(N^2).
  • Space Complexity: O(2^N N), the size of the output dominating the final space complexity.
    public List<List<Integer>> allPathsSourceTarget(int[][] graph) {
        return solve(graph, 0);
    }

    public List<List<Integer>> solve(int[][] graph, int node) {
        int N = graph.length;
        List<List<Integer>> ans = new ArrayList();
        if (node == N - 1) {
            List<Integer> path = new ArrayList();
            path.add(N-1);
            ans.add(path);
            return ans;
        }

        for (int nei: graph[node]) {
            for (List<Integer> path: solve(graph, nei)) {
                path.add(0, node);
                ans.add(path);
            }
        }
        return ans;
    }

http://zxi.mytechroad.com/blog/graph/leetcode-797-all-paths-from-source-to-target/
  vector<vector<int>> allPathsSourceTarget(vector<vector<int>>& graph) {    
    vector<vector<int>> ans;
    vector<int> path{0};    
    dfs(graph, path, ans);
    return ans;
  }
private:
  void dfs(const vector<vector<int>>& graph,
           vector<int>& path, vector<vector<int>>& ans) {
    if (path.back() == graph.size() - 1) {
      ans.push_back(path);
      return;
    }
    
    for (int next : graph[path.back()]) {
      path.push_back(next);
      dfs(graph, path, ans);
      path.pop_back();
    }
  }


// TLE, but after remove useCache, it passed
// need remove duplicate paths
public List<List<Integer>> allPathsSourceTarget(int[][] graph) {
if (graph == null)
return new ArrayList<>();
Set<LinkedHashSet<Integer>> allPaths = new HashSet<>();
allPathsSourceTarget(graph, 0, new LinkedHashSet<>(), allPaths);

return allPaths.stream().map(el -> new ArrayList<>(el)).collect(Collectors.toList());
}

private void allPathsSourceTarget(int[][] graph, int node, LinkedHashSet<Integer> path,
Set<LinkedHashSet<Integer>> allPaths) {
path.add(node); // path.add(node) should be here
if (node == graph.length - 1) {
allPaths.add(new LinkedHashSet<>(path)); // mistake: allPaths.add(path);
path.remove(node); // don't miss this
return;
}

// cache
if (useCache(allPaths, node, path)) {
path.remove(node);
return;
}

for (int i = 0; i < graph[node].length; i++) {
allPathsSourceTarget(graph, graph[node][i], path, allPaths);
}

path.remove(node);
}

private boolean useCache(Set<LinkedHashSet<Integer>> allPaths, int node, LinkedHashSet<Integer> path) {
boolean nodeVisited = false;
List<LinkedHashSet<Integer>> newAllPaths = new ArrayList<>();

for (LinkedHashSet<Integer> cache : allPaths) {
if (cache.contains(node)) {
nodeVisited = true;
LinkedHashSet<Integer> newPath = new LinkedHashSet<>(path);
boolean foundIt = false;
for (Integer tmp : cache) {
if (foundIt) {
newPath.add(tmp);
} else if (Objects.equals(tmp, node)) {
foundIt = true;
} // else continue
}

// ConcurrentModificationException
// allPaths.add(newPath);

newAllPaths.add(newPath);// don't miss this
}
}

allPaths.addAll(newAllPaths);
return nodeVisited;

}



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