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

November LeetCoding Challenge 28

Description

All Paths From Source to Target

Given a directed acyclic graph (DAG) of n nodes labeled from 0 to n - 1, find all possible paths from node 0 to node n - 1 and return them in any order.

The graph is given as follows: graph[i] is a list of all nodes you can visit from node i (i.e., there is a directed edge from node i to node graph[i][j]).

Example 1:

Input: graph = [[1,2],[3],[3],[]]
Output: [[0,1,3],[0,2,3]]
Explanation: There are two paths: 0 -> 1 -> 3 and 0 -> 2 -> 3.

Example 2:

Input: graph = [[4,3,1],[3,2,4],[3],[4],[]]
Output: [[0,4],[0,3,4],[0,1,3,4],[0,1,2,3,4],[0,1,4]]

Example 3:

Input: graph = [[1],[]]
Output: [[0,1]]

Example 4:

Input: graph = [[1,2,3],[2],[3],[]]
Output: [[0,1,2,3],[0,2,3],[0,3]]

Example 5:

Input: graph = [[1,3],[2],[3],[]]
Output: [[0,1,2,3],[0,3]]

Constraints:

• n == graph.length
• 2 <= n <= 15
• 0 <= graph[i][j] < n
• graph[i][j] != i (i.e., there will be no self-loops).
• All the elements of graph[i] are unique.
• The input graph is guaranteed to be a DAG.

Solution

class Solution {
public:
  vector<vector<int>> allPathsSourceTarget(vector<vector<int>>& graph) {
    int degree[15] = {};
    int n = graph.size();
    for(auto &nexts : graph) {
      for(auto next : nexts) {
        degree[next] += 1;
      }
    }
    if(degree[0]) return {};

    queue<int> q;
    vector<vector<int>> result[n];
    q.push(0);
    result[0].push_back({0});

    while(q.size()) {
      int cur = q.front();
      q.pop();
      for(auto next : graph[cur]) {
        for(auto path : result[cur]) {
          path.push_back(next);
          result[next].emplace_back(path);
        }
        degree[next] -= 1;
        if(!degree[next]) {
          q.push(next);
        }
      }
    }
    return result[n - 1];
  }
};

// Accepted
// 30/30 cases passed (28 ms)
// Your runtime beats 33.13 % of cpp submissions
// Your memory usage beats 15.98 % of cpp submissions (18.2 MB)