- Difficulty: Medium
- Tags: LeetCode, Medium, Array, Binary Search, Sorting, leetcode-2070, Sort, O(nlogn + qlogn), O(1)
Problem
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