Most Beautiful Item for Each Query

lcpy gen -n 2070   # by problem number
lcpy gen -s most_beautiful_item_for_each_query   # by problem name

Input: items = [[1,2],[3,2],[2,4],[5,6],[3,5]], queries = [1,2,3,4,5,6]
Output: [2,4,5,5,6,6]
Explanation:
- For queries[0]=1, [1,2] is the only item which has price <= 1. Hence, the answer for this query is 2.
- For queries[1]=2, the items which can be considered are [1,2] and [2,4]. The maximum beauty among them is 4.
- For queries[2]=3 and queries[3]=4, the items which can be considered are [1,2], [3,2], [2,4], and [3,5]. The maximum beauty among them is 5.
- For queries[4]=5 and queries[5]=6, all items can be considered. Hence, the answer for them is the maximum beauty of all items, i.e., 6.

Input: items = [[1,2],[1,2],[1,3],[1,4]], queries = [1]
Output: [4]
Explanation: The price of every item is equal to 1, so we choose the item with the maximum beauty 4. Note that multiple items can have the same price and/or beauty.

Input: items = [[10,1000]], queries = [5]
Output: [0]
Explanation: No item has a price less than or equal to 5, so no item can be chosen. Hence, the answer to the query is 0.

from bisect import bisect_right


class Solution:
    # Time: O(n log n + m log n) for n items and m queries
    # Space: O(n)
    def maximum_beauty(self, items: list[list[int]], queries: list[int]) -> list[int]:
        items = sorted(items)
        prices: list[int] = []
        best: list[int] = []
        max_beauty = 0
        for price, beauty in items:
            max_beauty = max(max_beauty, beauty)
            prices.append(price)
            best.append(max_beauty)
        result: list[int] = []
        for query in queries:
            i = bisect_right(prices, query)
            result.append(best[i - 1] if i > 0 else 0)
        return result