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Leetcode 1431. Kids With the Greatest Number of Candies

2023-03-04 12:47 作者:您是打尖儿还是住店呢  | 我要投稿

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] <= 10

    Easy 题目,直接遍历一下即可;

Runtime: 1 ms, faster than 91.92% of Java online submissions for Kids With the Greatest Number of Candies.

Memory Usage: 42.5 MB, less than 32.91% of Java online submissions for Kids With the Greatest Number of Candies.


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