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My first instinct was just linear scan and I almost coded it up before catching myself.
Use binary search to find the first word that is >= each prefix, then check if it starts with the prefix. This leverages the sorted order to achieve O(m log n) time, where m is the number of prefixes and n is the number of words. Discuss the trade-offs of preprocessing (e.g., trie) versus per-query binary search.
Pro tip: Mention that you can optimize by using the built-in bisect module in Python or similar functions in other languages, and highlight that binary search is preferred when the list is static and queries are many. Also, note that if the list is huge and queries are frequent, a trie might be better despite higher memory usage.
Confirm that the list is sorted lexicographically, prefixes are non-empty strings, and the list may contain duplicates. Discuss handling of empty list or no match.
Decide between binary search per prefix (O(m log n)) and building a trie (O(n * L + m * L)). Justify based on constraints like list size, number of queries, and memory.
For each prefix, use binary search to find the leftmost index where the word is >= prefix. Check if that word starts with the prefix; if yes, return index, else -1.
State that binary search takes O(log n) per prefix, so O(m log n) total, with O(1) extra space. Compare with trie: O(n * L) preprocessing and O(L) per query, but O(n * L) space.
Mention that if the list is static and queries are many, binary search is simple and efficient. If queries are frequent and memory allows, a trie can be faster. Also, note that sorting the prefixes or using a batch approach might help.
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