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Graph Question - Medium Level - Question 2

 2192. All Ancestors of a Node in a Directed Acyclic Graph

You are given a positive integer n representing the number of nodes of a Directed Acyclic Graph (DAG). The nodes are numbered from 0 to n - 1 (inclusive).

You are also given a 2D integer array edges, where edges[i] = [fromi, toi] denotes that there is a unidirectional edge from fromi to toi in the graph.

Return a list answer, where answer[i] is the list of ancestors of the ith node, sorted in ascending order.

A node u is an ancestor of another node v if u can reach v via a set of edges.


Constraints:

1 <= n <= 1000

0 <= edges.length <= min(2000, n * (n - 1) / 2)

edges[i].length == 2

0 <= fromi, toi <= n - 1

fromi != toi

There are no duplicate edges.

The graph is directed and acyclic.


Analysis:


This question can be solved by constructing a graph by the edges information.

Since this question asks for the parents information, we can create a graph storing the direct parents notes. 

After having the graph, we can do a top-down dfs, to search through each sub-tree, and merge the results from all sub-trees into one vector which are the all the ancestors of the current node.

To reduce the redundant searching, a visited vector can be used to record the visiting status. Once visited once, the status will be labeled as 1, then all the following visits can return the previous result directly without searching again.

So the method can be viewed as dfs + dp in general.


Time complexity: for each node, we perform one dfs. For each dfs, we can reuse the results from previous dfs with the visited vector (dp), so the averaged time complexity for this step is O(N), where N is the number of nodes. But there is still another step in each dfs, merging ancestors into one vector, which is O(Nlog(N)) in average. So the overall time complexity is O(N^2*log(N)).

The space time complexity is O(N^2).


See the code below:


class Solution {
public:
    vector<vector<int>> getAncestors(int n, vector<vector<int>>& edges) {
        vector<vector<int>> res(n), g(n);
        vector<int> visited(n, 0);
        for(auto &a : edges) g[a[1]].push_back(a[0]);
        for(int i=0; i<n; ++i) {
            dfs(i, g, visited, res);
        }
        return res;
    }
private:
    vector<int> dfs(int i, vector<vector<int>>& g, vector<int>& vit, vector<vector<int>>& res) {
        if(vit[i]) {
            return res[i];
        }
        vit[i] = 1;
        vector<int> ret;
        for(auto &a : g[i]) {
            vector<int> one = dfs(a, g, vit, res);
            for(auto &b : one) ret.push_back(b);
            ret.push_back(a);
        }
        sort(ret.begin(), ret.end());
        vector<int> copy;
        for(auto &a : ret) {
            if(copy.empty() || copy.back() != a) copy.push_back(a);
        }
        return res[i] = copy;
    }
};


Upper Layer

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