Leetcode: Sequence Reconstruction

Check whether the original sequence org can be uniquely reconstructed from the sequences in seqs. The org sequence is a permutation of the integers from 1 to n, with 1 ≤ n ≤ 104. Reconstruction means building a shortest common supersequence of the sequences in seqs (i.e., a shortest sequence so that all sequences in seqs are subsequences of it). Determine whether there is only one sequence that can be reconstructed from seqs and it is the org sequence.

Example 1:

Input:
org: [1,2,3], seqs: [[1,2],[1,3]]

Output:
false

Explanation:
[1,2,3] is not the only one sequence that can be reconstructed, because [1,3,2] is also a valid sequence that can be reconstructed.
Example 2:

Input:
org: [1,2,3], seqs: [[1,2]]

Output:
false

Explanation:
The reconstructed sequence can only be [1,2].
Example 3:

Input:
org: [1,2,3], seqs: [[1,2],[1,3],[2,3]]

Output:
true

Explanation:
The sequences [1,2], [1,3], and [2,3] can uniquely reconstruct the original sequence [1,2,3].
Example 4:

Input:
org: [4,1,5,2,6,3], seqs: [[5,2,6,3],[4,1,5,2]]

Output:
true

Topological Sort: This problem is to determine if there‘s one, and only one sequence to sort a DAG. The method is to check if the queue‘s size is always 1 or not. If the queue has over 1 size when we‘re conducting topological sort, we return false, which implies that there exists more than 1 sequence to sort this DAG

Some corner case that i missed when write it:

Input:[1] [[1],[2,3],[3,2]]

Output:true

Expected:false

How to revise: check if index at last equals graph‘s size

 1 public class Solution {
 2     public boolean sequenceReconstruction(int[] org, int[][] seqs) {
 3         HashMap<Integer, HashSet<Integer>> graph = new HashMap<>();
 4         HashMap<Integer, Integer> indegree = new HashMap<>();
 5
 6         //build the graph
 7         for (int[] seq : seqs) {
 8             if (seq.length == 1) {
 9                 if (!graph.containsKey(seq[0])) {
10                     graph.put(seq[0], new HashSet<Integer>());
11                     indegree.put(seq[0], 0);
12                 }
13             }
14             else {
15                 for (int i=0; i<seq.length-1; i++) {
16                     if (!graph.containsKey(seq[i])) {
17                         graph.put(seq[i], new HashSet<Integer>());
18                         indegree.put(seq[i], 0);
19                     }
20                     if (!graph.containsKey(seq[i+1])) {
21                         graph.put(seq[i+1], new HashSet<Integer>());
22                         indegree.put(seq[i+1], 0);
23                     }
24                     if (!graph.get(seq[i]).contains(seq[i+1])) {
25                         graph.get(seq[i]).add(seq[i+1]);
26                         indegree.put(seq[i+1], indegree.get(seq[i+1])+1);
27                     }
28                 }
29             }
30         }
31
32         //Topological sort, if any time the BFS queue‘s size > 1, return false;
33         Queue<Integer> queue = new LinkedList<>();
34         for (Map.Entry<Integer, Integer> entry : indegree.entrySet()) {
35             if (entry.getValue() == 0) {
36                 queue.offer(entry.getKey());
37             }
38         }
39
40         int index = 0;
41         while (!queue.isEmpty()) {
42             int size = queue.size();
43             if (size > 1) return false;
44             int cur = queue.poll();
45             if (index>=org.length || org[index++] != cur) return false;
46             HashSet<Integer> neighbors = graph.get(cur);
47             for (int neighbor : neighbors) {
48                 indegree.put(neighbor, indegree.get(neighbor)-1);
49                 if (indegree.get(neighbor) == 0) {
50                     queue.offer(neighbor);
51                 }
52             }
53         }
54         return (index==org.length)&&(index==indegree.size())? true : false;
55     }
56 }