LeetCode 444. 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


  • 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.

Python Solution

Check if the result of topological order of sequences is the same as the given order.

class Solution:
    def sequenceReconstruction(self, org: List[int], seqs: List[List[int]]) -> bool:
        graph = self.build_graph(seqs)
        topological_order = self.topological_sort(graph)
        return topological_order == org        
    def build_graph(self, seqs):        
        graph = {}
        for seq in seqs:
            for node in seq:
                if node not in graph:
                    graph[node] = set()
        for seq in seqs:
            for i in range(1, len(seq)):
                graph[seq[i - 1]].add(seq[i])

        return graph
    def topological_sort(self, graph):
        indegrees = {
            node: 0 for node in graph
        for node in graph:
            for neighbor in graph[node]:
                indegrees[neighbor] += 1
        topological_order = []
        queue = deque()
        for node in graph:
            if indegrees[node] == 0:
        while queue:
            if len(queue) > 1:
                return None
            node = queue.popleft()
            for neighbor in graph[node]:
                indegrees[neighbor] -= 1
                if indegrees[neighbor] == 0:
        if len(topological_order) == len(graph):
            return topological_order
        return None
  • Time Complexity: O(N).
  • Space Complexity: O(N).

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