Sequence Reconstruction

• Lintcode : 605 / Leetcode : 444
• Level : Medium

Problem

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 ≤ 10^4. 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
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

Concept & Algorithm

筆記

Time Complexity & Space Complexity

time : O(n)
space: O(n)

Answer

class Solution {
public:
    bool sequenceReconstruction(vector<int> &org, vector<vector<int>> &seqs) {
        unordered_map<int, unordered_multiset<int>> edges;
        unordered_map<int, int> degree;
        for (auto seq : seqs) {
            for (int i = 0; i < seq.size(); i++) {
                degree[seq[i]];
                if (i + 1 < seq.size()) {
                    edges[seq[i]].insert(seq[i + 1]);
                    degree[seq[i + 1]]++;
                }
            }
        }

        queue<int> q;
        vector<int> ans;
        for (auto it = degree.begin(); it != degree.end(); it++) {
            if (it -> second == 0) q.push(it -> first);  
        }
        while (!q.empty()) {
            if (q.size() > 1) return false;
            int h = q.front(); q.pop();
            ans.push_back(h);
            for (auto it = edges[h].begin(); it != edges[h].end(); it++) {
                degree[*it]--;
                if (degree[*it] == 0) q.push(*it);
            }
        }
        if (org.size() != ans.size()) return false;
        for (int i = 0; i < org.size(); i++) {
            if (org[i] != ans[i]) return false;
        }
        return true;
    }
};

tags: topological sorting

​