Today I have done leetcode's June LeetCoding Challenge with cpp.

June LeetCoding Challenge 29

Description

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.

Example 1:

Input: n = 8, edgeList = [[0,3],[0,4],[1,3],[2,4],[2,7],[3,5],[3,6],[3,7],[4,6]]
Output: [[],[],[],[0,1],[0,2],[0,1,3],[0,1,2,3,4],[0,1,2,3]]
Explanation:
The above diagram represents the input graph.
- Nodes 0, 1, and 2 do not have any ancestors.
- Node 3 has two ancestors 0 and 1.
- Node 4 has two ancestors 0 and 2.
- Node 5 has three ancestors 0, 1, and 3.
- Node 6 has five ancestors 0, 1, 2, 3, and 4.
- Node 7 has four ancestors 0, 1, 2, and 3.

Example 2:

Input: n = 5, edgeList = [[0,1],[0,2],[0,3],[0,4],[1,2],[1,3],[1,4],[2,3],[2,4],[3,4]]
Output: [[],[0],[0,1],[0,1,2],[0,1,2,3]]
Explanation:
The above diagram represents the input graph.
- Node 0 does not have any ancestor.
- Node 1 has one ancestor 0.
- Node 2 has two ancestors 0 and 1.
- Node 3 has three ancestors 0, 1, and 2.
- Node 4 has four ancestors 0, 1, 2, and 3.

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.

Solution

class Solution {
  void dfs(
    int parent,
    int node,
    vector<bool> &visit,
    const vector<vector<int>> &graph,
    vector<vector<int>> &answer
  ) {
    visit[node] = true;
    for(auto next : graph[node]) {
      if(visit[next]) continue;
      answer[next].push_back(parent);
      dfs(parent, next, visit, graph, answer);
    }
  }
public:
  vector<vector<int>> getAncestors(int n, vector<vector<int>>& edges) {
    vector<vector<int>> graph(n);
    vector<vector<int>> answer(n);

    for(auto & edge : edges) {
      graph[edge[0]].push_back(edge[1]);
    }
    for(int i = 0; i < n; ++i) {
      vector<bool> visit(n);
      dfs(i, i, visit, graph, answer);
    }

    for(auto &result : answer) {
      sort(result.begin(), result.end());
      result.resize(unique(result.begin(), result.end()) - result.begin());
    }

    return answer;
  }
};