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My first instinct was to sort, which was wrong.
Model the fragments as a directed graph where each fragment is an edge from start ID to end ID, then find the Eulerian path that visits every edge exactly once. Since the fragments form a single chain, you can also use a hash map from start ID to fragment and traverse from the unique start node (one with no incoming edges) to reconstruct the sequence in O(n) time.
Pro tip: Clarify whether the chain is guaranteed to be linear and complete; if so, the hash map approach is simpler and faster than a full Eulerian path algorithm. Mention that you would validate the input to handle edge cases like duplicate start IDs or cycles.
Confirm that fragments form a single chain with no branching, and that each fragment's end ID matches the next fragment's start ID. Identify the unique start fragment (start ID not appearing as any end ID) and end fragment (end ID not appearing as any start ID).
Use a hash map (dictionary) to map each start ID to its corresponding fragment object, enabling O(1) lookup. Alternatively, build an adjacency list if the graph might have branches, but for a single chain, the hash map is sufficient.
Start from the unique start fragment, append its payload to the result, then follow the chain by looking up the next fragment using the current fragment's end ID. Continue until no further fragment exists.
Check for empty input, single fragment, or invalid chains (e.g., missing links, cycles). If the chain is not guaranteed, consider using an Eulerian path algorithm to handle general cases.
State that the time complexity is O(n) for building the map and traversing the chain, and space complexity is O(n). Walk through a small example to verify correctness.
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