Skip to main content

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

Comments

Post a Comment

Popular posts from this blog

Segment Tree

Segment tree can be viewed as an abstract data structure which using some more space to trade for speed. For example, for a typical question with O(N^2) time complexity, the segment tree method can decrease it to O(N*log(N)).  To make it understandable, let us consider one example. Say we have an integer array of N size, and what we want is to query the maximum with a query range [idx1, idx2], where idx1 is the left indexes, and idx2 is the right indexes inclusive. If we only do this kind of query once, then we just need to scan through the array from idx1 to idx2 once, and record the maximum, done. The time complexity is O(N), which is decent enough in most cases even though it is not the optimal one (for example, with a segment tree built, the time complexity can decrease down to O(log(N))). However, how about we need to query the array N times? If we continue to use the naïve way above, then the time complexity is O(N^2), since for each query we need to scan the query range once...
Post a Comment
Read more

Bit Manipulation - Example

  Leetcode 136 Single Number Given a non-empty array of integers nums, every element appears twice except for one. Find that single one. You must implement a solution with a linear runtime complexity and use only constant extra space. Constraints: 1 <= nums.length <= 3 * 10^4 -3 * 10^4 <= nums[i] <= 3 * 10^4 Each element in the array appears twice except for one element which appears only once. Analysis: If there is no space limitation, this question can be solved by counting easily. But counting requires additional space. Here we can use xor (^) operation based on some interesting observations:  A^A = 0, here A is any number A^0 = A, here A is any number Since all the number appears twice except one, then all the number appear even numbers will be cancelled out, and only the number appears one time is left, which is what we want. See the code below: class Solution { public: int singleNumber(vector<int>& nums) { int res = 0; for(auto ...
Post a Comment
Read more

Rolling Hash

Rolling hash is one common trick used to increase efficiency of substring comparisons by compressing (or hashing) a string into a integer. After this step, we can compare two strings directly without comparing each chars. So the efficiency can be increased from O(N) to O(1). So how to implement the rolling hash? First we need to choose a base for the expansion and a modulo to mod. The basic formula is (suppose the window is n, and the rolling direction is from left to right), HashVal = (A1*p^(n-1) + A2*p^(n-2) + ... + An-1*p^1 + An*p^0)%mod where HashVal is the hash value, Ai is the ith element, p is the base, and mod is the modulo. To avoid collision as much as we can, p and modulo usually need to be large prime numbers. One corner case is that the base order in the above formula cannot be reversed. Or to be more clear, if the rolling direction is from left to right in an array, the first element should be in the highest order of the base, or times p^(n-1), and the last element times ...
Post a Comment
Read more