LeetCode 444: Sequence Reconstruction Solution

Master LeetCode problem 444 (Sequence Reconstruction), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

444. Sequence Reconstruction

Medium

Array Graph Topological Sort

Problem Explanation

Explanation:

The problem asks us to determine whether a given sequence org can be reconstructed from a set of sequences seqs. In order to reconstruct the sequence, the following conditions must be satisfied:

The sequence org must be a permutation of the integers from 1 to n, where n is the number of elements in org.
The sequences in seqs must maintain their relative order in org.

To solve this problem, we can perform the following steps:

Create a directed graph where each number in org is a node.
Add edges to the graph based on the sequences in seqs.
Perform a topological sort on the graph and check if the resulting order matches org.

: :

Solution Code

import java.util.*;

class Solution {
    public boolean sequenceReconstruction(int[] org, List<List<Integer>> seqs) {
        Map<Integer, Set<Integer>> graph = new HashMap<>();
        Map<Integer, Integer> indegree = new HashMap<>();

        for (List<Integer> seq : seqs) {
            for (int i = 0; i < seq.size(); i++) {
                graph.putIfAbsent(seq.get(i), new HashSet<>());
                indegree.putIfAbsent(seq.get(i), 0);

                if (i > 0) {
                    if (graph.get(seq.get(i - 1)).add(seq.get(i))) {
                        indegree.put(seq.get(i), indegree.get(seq.get(i)) + 1);
                    }
                }
            }
        }

        if (org.length != indegree.size()) {
            return false;
        }

        Queue<Integer> queue = new LinkedList<>();
        for (int num : indegree.keySet()) {
            if (indegree.get(num) == 0) {
                queue.offer(num);
            }
        }

        int index = 0;
        while (!queue.isEmpty()) {
            if (queue.size() > 1) {
                return false;
            }

            int current = queue.poll();
            if (org[index++] != current) {
                return false;
            }

            for (int neighbor : graph.get(current)) {
                indegree.put(neighbor, indegree.get(neighbor) - 1);
                if (indegree.get(neighbor) == 0) {
                    queue.offer(neighbor);
                }
            }
        }

        return index == org.length;
    }
}

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