2179 Most Beautiful Item For Each Query

Most Beautiful Item for Each Query

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

from typing import List

class Solution:
    def maximumBeauty(self, items: List[List[int]], queries: List[int]) -> List[int]:
        # Sort items by their price in ascending order, then by beauty in descending order
        items.sort()
        
        # Precompute the maximum beauty up to each price
        max_beauty = []
        current_max_beauty = 0
        for price, beauty in items:
            current_max_beauty = max(current_max_beauty, beauty)
            max_beauty.append((price, current_max_beauty))
        
        # Function to find the maximum beauty within the price limit using binary search
        def find_max_beauty(price_limit: int) -> int:
            left, right = 0, len(max_beauty) - 1
            while left <= right:
                mid = left + (right - left) // 2
                if max_beauty[mid][0] <= price_limit:
                    left = mid + 1
                else:
                    right = mid - 1
            return max_beauty[right][1] if right >= 0 else 0
        
        # Answer each query using binary search on max_beauty
        ans = []
        for query in queries:
            ans.append(find_max_beauty(query))
        
        return ans