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2070. Most Beautiful Item for Each Query

Difficulty Topics

Description

You are given a 2D integer array items where items[i] = [pricei, beautyi] denotes the price and beauty of an item respectively.

You are also given a 0-indexed integer array queries. For each queries[j], you want to determine the maximum beauty of an item whose price is less than or equal to queries[j]. If no such item exists, then the answer to this query is 0.

Return an array answer of the same length as queries where answer[j] is the answer to the jth query.

 

Example 1:

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.

Example 2:

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.

Example 3:

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.

 

Constraints:

  • 1 <= items.length, queries.length <= 105
  • items[i].length == 2
  • 1 <= pricei, beautyi, queries[j] <= 109

Solution

most-beautiful-item-for-each-query.py
class Solution:
    def maximumBeauty(self, items: List[List[int]], queries: List[int]) -> List[int]:
        n = len(items)
        items.sort(key = lambda x : (x[0], -x[1]))
        res = []
        A = [(0, 0)]
        curr = 0

        for i, (price, beauty) in enumerate(items):
            if i > 1 and price == items[i - 1][0]: continue

            curr = max(curr, beauty)
            A.append((price, curr))

        for q in queries:
            index = bisect.bisect(A, (q + 1, )) - 1
            res.append(A[index][1])

        return res