README_EN.md

444. Sequence Reconstruction

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Description

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

 

Constraints:

• 1 <= n <= 10^4
• org is a permutation of {1,2,...,n}.
• 1 <= segs[i].length <= 10^5
• seqs[i][j] fits in a 32-bit signed integer.

 

UPDATE (2017/1/8):
The seqs parameter had been changed to a list of list of strings (instead of a 2d array of strings). Please reload the code definition to get the latest changes.

Solutions

Python3

class Solution:
    def sequenceReconstruction(self, org: List[int], seqs: List[List[int]]) -> bool:
        n = len(org)
        nums = set()
        for seq in seqs:
            for num in seq:
                if num < 1 or num > n:
                    return False
                nums.add(num)
        if len(nums) < n:
            return False

        edges = collections.defaultdict(list)
        indegree = [0] * (n + 1)
        for seq in seqs:
            i = seq[0]
            for j in seq[1:]:
                edges[i].append(j)
                indegree[j] += 1
                i = j
        q = collections.deque()
        for i in range(1, n + 1):
            if indegree[i] == 0:
                q.append(i)
        cnt = 0
        while q:
            if len(q) > 1 or org[cnt] != q[0]:
                return False
            i = q.popleft()
            cnt += 1
            for j in edges[i]:
                indegree[j] -= 1
                if indegree[j] == 0:
                    q.append(j)
        return cnt == n

Java

class Solution {
    public boolean sequenceReconstruction(int[] org, List<List<Integer>> seqs) {
        int n = org.length;
        Set<Integer> nums = new HashSet<>();
        for (List<Integer> seq : seqs) {
            for (int num : seq) {
                if (num < 1 || num > n) {
                    return false;
                }
                nums.add(num);
            }
        }
        if (nums.size() < n) {
            return false;
        }
        List<Integer>[] edges = new List[n + 1];
        for (int i = 0; i < edges.length; ++i) {
            edges[i] = new ArrayList<>();
        }
        int[] indegree = new int[n + 1];
        for (List<Integer> seq : seqs) {
            int i = seq.get(0);
            for (int j = 1; j < seq.size(); ++j) {
                edges[i].add(seq.get(j));
                ++indegree[seq.get(j)];
                i = seq.get(j);
            }
        }
        Queue<Integer> q = new LinkedList<>();
        for (int i = 1; i <= n; ++i) {
            if (indegree[i] == 0) {
                q.offer(i);
            }
        }
        int cnt = 0;
        while (!q.isEmpty()) {
            if (q.size() > 1 || q.peek() != org[cnt]) {
                return false;
            }
            ++cnt;
            int i = q.poll();
            for (int j : edges[i]) {
                --indegree[j];
                if (indegree[j] == 0) {
                    q.offer(j);
                }
            }
        }
        return cnt == n;
    }
}

C++

class Solution {
public:
    bool sequenceReconstruction(vector<int>& org, vector<vector<int>>& seqs) {
        int n = org.size();
        unordered_set<int> nums;
        for (auto& seq : seqs)
        {
            for (int num : seq)
            {
                if (num < 1 || num > n) return false;
                nums.insert(num);
            }
        }
        if (nums.size() < n) return false;
        vector<vector<int>> edges(n + 1);
        vector<int> indegree(n + 1);
        for (auto& seq : seqs)
        {
            int i = seq[0];
            for (int j = 1; j < seq.size(); ++j)
            {
                edges[i].push_back(seq[j]);
                ++indegree[seq[j]];
                i = seq[j];
            }
        }
        queue<int> q;
        for (int i = 1; i <= n; ++i)
        {
            if (indegree[i] == 0) q.push(i);
        }
        int cnt = 0;
        while (!q.empty())
        {
            if (q.size() > 1 || q.front() != org[cnt]) return false;
            ++cnt;
            int i = q.front();
            q.pop();
            for (int j : edges[i])
            {
                --indegree[j];
                if (indegree[j] == 0) q.push(j);
            }
        }
        return cnt == n;
    }
};

Go

func sequenceReconstruction(org []int, seqs [][]int) bool {
	n := len(org)
	nums := make(map[int]bool)
	for _, seq := range seqs {
		for _, num := range seq {
			if num < 1 || num > n {
				return false
			}
			nums[num] = true
		}
	}
	if len(nums) < n {
		return false
	}
	edges := make([][]int, n+1)
	indegree := make([]int, n+1)
	for _, seq := range seqs {
		i := seq[0]
		for _, j := range seq[1:] {
			edges[i] = append(edges[i], j)
			indegree[j]++
			i = j
		}
	}
	var q []int
	for i := 1; i <= n; i++ {
		if indegree[i] == 0 {
			q = append(q, i)
		}
	}
	cnt := 0
	for len(q) > 0 {
		if len(q) > 1 || org[cnt] != q[0] {
			return false
		}
		i := q[0]
		q = q[1:]
		cnt++
		for _, j := range edges[i] {
			indegree[j]--
			if indegree[j] == 0 {
				q = append(q, j)
			}
		}
	}
	return cnt == n
}

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