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Model the fragments as a linked list where each fragment points to the next via its end tag. Use a hash map to map start tags to fragments, then find the head (the start tag that is not any fragment's end tag) and traverse the chain, concatenating payloads. This yields O(n) time and space.
Pro tip: Clarify assumptions upfront: tags are unique, no cycles, and exactly one chain. Mention that you can also build the chain by linking end tags to start tags, but finding the head is key to avoid extra traversal.
Confirm that fragments form exactly one continuous chain, tags are unique, and there are no cycles. Ask about input size and whether payloads can be empty.
Use a hash map to map each start tag to its fragment for O(1) lookup. Use a set to track all end tags to identify the head.
Iterate through fragments and collect all end tags. The head is the fragment whose start tag is not in the end tags set.
Starting from the head, repeatedly look up the next fragment by the current fragment's end tag, appending its payload, until no next fragment exists.
State that the algorithm runs in O(n) time and uses O(n) space, where n is the number of fragments. Discuss potential optimizations if needed.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
This tripped me up more than it should have.
First, clarify the problem: you're given a chain of fragments where each fragment has two payloads, and you need to find an ordering that forms a valid chain, considering both forward and backward traversal. Then, model the fragments as edges in an undirected graph and find an Eulerian path, ensuring you handle both orientations of each fragment and output either ordering of payloads as valid.
Pro tip: Mention that you can treat each fragment as an undirected edge and use Hierholzer's algorithm to find an Eulerian path in O(E) time, which is optimal. Also, discuss how to handle disconnected components and verify that the graph has 0 or 2 odd-degree vertices.
Restate the problem to ensure understanding: fragments are bidirectional, and either ordering of payloads is acceptable. Ask about input size, whether fragments can be reused, and if the chain must use all fragments.
Represent each fragment as an undirected edge between its two payloads. The goal is to find an Eulerian path that uses every edge exactly once, which corresponds to a valid chain.
Verify that the graph is connected (ignoring isolated vertices) and has either 0 or 2 vertices with odd degree. If not, no valid chain exists.
Use Hierholzer's algorithm: start from an odd-degree vertex if one exists, otherwise any vertex with edges, and perform a DFS to construct the path by splicing cycles.
Convert the Eulerian path back to a sequence of fragments, ensuring each consecutive pair shares a payload. Output the chain in either direction as required.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Clarify that the problem involves assembling multiple independent chains from fragments, where each chain has no branching. Propose a graph-based solution: treat each fragment as a node and directed edges for overlaps, then find connected components and topologically sort each to reconstruct the chains.
Pro tip: Mention that you would validate the input to ensure each fragment has at most one predecessor and successor, and handle edge cases like cycles or isolated fragments. This shows attention to robustness and real-world data issues.
Restate the problem: given fragments that form multiple disjoint chains, assemble each chain into a single string. Confirm that each chain is linear (no branching) and that fragments may overlap.
Represent each fragment as a node. For every pair of fragments, if the suffix of one overlaps with the prefix of another, add a directed edge. This creates a directed graph where each node has at most one outgoing and one incoming edge (due to no branching).
Use union-find or DFS to find connected components (ignoring edge direction). Each component corresponds to one chain. Ensure that within each component, the graph is a simple path (no cycles, no branching).
For each component, find the start node (no incoming edge) and traverse the path, merging overlapping fragments. If a cycle is detected, handle it as an error or special case.
Collect the merged strings from each component and return them as a list. Discuss time complexity: O(n^2) for overlap checks, but can be optimized with a trie or hash map.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.