Question

444. Sequence Reconstruction

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Description

You are given an integer array nums of length n where nums is a permutation of the integers in the range [1, n]. You are also given a 2D integer array sequences where sequences[i] is a subsequence of nums.

Check if nums is the shortest possible and the only supersequence. The shortest supersequence is a sequence with the shortest length and has all sequences[i] as subsequences. There could be multiple valid supersequences for the given array sequences.

• For example, for sequences = [[1,2],[1,3]], there are two shortest supersequences, [1,2,3] and [1,3,2].
• While for sequences = [[1,2],[1,3],[1,2,3]], the only shortest supersequence possible is [1,2,3]. [1,2,3,4] is a possible supersequence but not the shortest.

Return true_ if _nums_ is the only shortest supersequence for _sequences_, or _false_ otherwise_.

A subsequence is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements.

 

Example 1:

Input: nums = [1,2,3], sequences = [[1,2],[1,3]]
Output: false
Explanation: There are two possible supersequences: [1,2,3] and [1,3,2].
The sequence [1,2] is a subsequence of both: [1,2,3] and [1,3,2].
The sequence [1,3] is a subsequence of both: [1,2,3] and [1,3,2].
Since nums is not the only shortest supersequence, we return false.

Example 2:

Input: nums = [1,2,3], sequences = [[1,2]]
Output: false
Explanation: The shortest possible supersequence is [1,2].
The sequence [1,2] is a subsequence of it: [1,2].
Since nums is not the shortest supersequence, we return false.

Example 3:

Input: nums = [1,2,3], sequences = [[1,2],[1,3],[2,3]]
Output: true
Explanation: The shortest possible supersequence is [1,2,3].
The sequence [1,2] is a subsequence of it: [1,2,3].
The sequence [1,3] is a subsequence of it: [1,2,3].
The sequence [2,3] is a subsequence of it: [1,2,3].
Since nums is the only shortest supersequence, we return true.

 

Constraints:

• n == nums.length
• 1 <= n <= 104
• nums is a permutation of all the integers in the range [1, n].
• 1 <= sequences.length <= 104
• 1 <= sequences[i].length <= 104
• 1 <= sum(sequences[i].length) <= 105
• 1 <= sequences[i][j] <= n
• All the arrays of sequences are unique.
• sequences[i] is a subsequence of nums.

Solutions

Solution 1

class Solution:
  def sequenceReconstruction(
    self, nums: List[int], sequences: List[List[int]]
  ) -> bool:
    g = defaultdict(list)
    indeg = [0] * len(nums)
    for seq in sequences:
      for a, b in pairwise(seq):
        g[a - 1].append(b - 1)
        indeg[b - 1] += 1
    q = deque(i for i, v in enumerate(indeg) if v == 0)
    while q:
      if len(q) > 1:
        return False
      i = q.popleft()
      for j in g[i]:
        indeg[j] -= 1
        if indeg[j] == 0:
          q.append(j)
    return True

class Solution {
  public boolean sequenceReconstruction(int[] nums, List<List<Integer>> sequences) {
    int n = nums.length;
    int[] indeg = new int[n];
    List<Integer>[] g = new List[n];
    Arrays.setAll(g, k -> new ArrayList<>());
    for (var seq : sequences) {
      for (int i = 1; i < seq.size(); ++i) {
        int a = seq.get(i - 1) - 1, b = seq.get(i) - 1;
        g[a].add(b);
        indeg[b]++;
      }
    }
    Deque<Integer> q = new ArrayDeque<>();
    for (int i = 0; i < n; ++i) {
      if (indeg[i] == 0) {
        q.offer(i);
      }
    }
    while (!q.isEmpty()) {
      if (q.size() > 1) {
        return false;
      }
      int i = q.poll();
      for (int j : g[i]) {
        if (--indeg[j] == 0) {
          q.offer(j);
        }
      }
    }
    return true;
  }
}

class Solution {
public:
  bool sequenceReconstruction(vector<int>& nums, vector<vector<int>>& sequences) {
    int n = nums.size();
    vector<vector<int>> g(n);
    vector<int> indeg(n);
    for (auto& seq : sequences) {
      for (int i = 1; i < seq.size(); ++i) {
        int a = seq[i - 1] - 1, b = seq[i] - 1;
        g[a].push_back(b);
        ++indeg[b];
      }
    }
    queue<int> q;
    for (int i = 0; i < n; ++i)
      if (indeg[i] == 0) q.push(i);
    while (!q.empty()) {
      if (q.size() > 1) return false;
      int i = q.front();
      q.pop();
      for (int j : g[i])
        if (--indeg[j] == 0) q.push(j);
    }
    return true;
  }
};

func sequenceReconstruction(nums []int, sequences [][]int) bool {
  n := len(nums)
  g := make([][]int, n)
  indeg := make([]int, n)
  for _, seq := range sequences {
    for i := 1; i < len(seq); i++ {
      a, b := seq[i-1]-1, seq[i]-1
      g[a] = append(g[a], b)
      indeg[b]++
    }
  }
  q := []int{}
  for i, v := range indeg {
    if v == 0 {
      q = append(q, i)
    }
  }
  for len(q) > 0 {
    if len(q) > 1 {
      return false
    }
    i := q[0]
    q = q[1:]
    for _, j := range g[i] {
      indeg[j]--
      if indeg[j] == 0 {
        q = append(q, j)
      }
    }
  }
  return true
}