Class Solution
java.lang.Object
g2001_2100.s2070_most_beautiful_item_for_each_query.Solution
2070 - Most Beautiful Item for Each Query\.
Medium
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-
Constructor Summary
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Method Summary
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Solution
public Solution()
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Method Details
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maximumBeauty
public int[] maximumBeauty(int[][] items, int[] queries)
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