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 <= 10^4orgis a permutation of {1,2,...,n}.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.
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這題難點是理解題目，注意subsequence可以不連續，但是題目給的例子都是連續的，這點搞得非常困惑。。。如果不連續那這題的意思就清楚多了。。。
class Solution:
def sequenceReconstruction(self, org, seqs):
index = {num: i for i, num in enumerate([None] + org)}
pairs = set(zip([None] + org, org))
for seq in seqs:
for a, b in zip([None] + seq, seq):
if index[a] >= index.get(b, 0):
return False
pairs.discard((a, b))
return not pairs