Code Craft: Eligible Orders / Neighbor-Check Sweep

Code Craft round centered on processing an ordered list of orders where each removal can re-enable its neighbors. Naive `O(n²)` works but interviewer pushes for `O(n)` using a doubly-linked-list / neighbor-eligibility recomputation pattern. Requirements Input: a list of orders with some eligibility predicate (varies by interviewer — distance threshold, time-window adjacency, etc.). Repeatedly pick an eligible order, append it to the output, and remove it from the list. Removing an order can cause its neighbors to become newly eligible. Output: the order of removals. Notes Naive: scan the list, find an eligible order, remove, repeat. O(n²) because each scan touches the whole list. Optimized O(n): maintain a doubly-linked list over the orders and a set of currently eligible orders. When you remove an order, only check its (up to) two neighbors and possibly add them to the eligible set. Each order's eligibility is checked a bounded number of times. Sketching the linked-list + eligibility-set data structure clearly is the graded signal; the interviewer wants to see the abstraction, not just the optimization. Variants of this round appear with different eligibility predicates; the algorithmic skeleton (DLL + eligibility set + neighbor recompute on remove) carries over. The poster cited GPT-assisted explanation post-interview as the way they understood the optimal version — confirms the problem is non-trivial cold. Preparation Write a DLL-based remove-and-recompute pattern from memory; this is the core skill for the round. Practice articulating amortized analysis ("each neighbor is re-checked at most twice across the whole process, so total work is O(n)"). Have a small concrete example ready (5–6 elements) to walk through with the interviewer before coding.