The greedy insight is straightforward once you see it: sort products by price ascending (breaking ties by product ID), build a prefix sum array, then for each customer budget do a binary search to find how many you can afford from the front of that sorted list.
Sort products by price and precompute prefix sums to enable binary search for the maximum number of products within a budget. For each query, use binary search to find the maximum count, then handle tie-breaking by selecting the smallest total spend and lexicographically smallest product IDs using a greedy approach with a min-heap or sorted list.
Pro tip: Emphasize that the preprocessing step of sorting and prefix sums is crucial for achieving O(N log N) preprocessing and O(log N) per query, and clearly explain how tie-breaking is resolved without compromising the time complexity.
Sort products by price ascending and compute prefix sums of prices. This allows O(1) calculation of total cost for any number of cheapest products.
For a given budget, binary search on the prefix sums to find the maximum k such that the sum of the k cheapest products is ≤ budget. This gives the maximum number of distinct products.
If multiple sets of k products have the same total spend, choose the one with the smallest total spend (already ensured by using cheapest products). For lexicographically smallest product IDs, among all products with price ≤ the k-th cheapest price, select the smallest IDs greedily.
Preprocess a data structure (e.g., segment tree or sorted list with binary search) to quickly retrieve the lexicographically smallest set of k product IDs that fit the budget, ensuring O(log N) per query.
Argue that the greedy choice of cheapest products maximizes count, and the tie-breaking rules are satisfied. Analyze time: O(N log N) preprocessing (sorting) and O(log N) per query (binary search and data structure lookup).
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