README.md

2070. 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

Solutions (Rust)

1. Solution

impl Solution {
    pub fn maximum_beauty(items: Vec<Vec<i32>>, queries: Vec<i32>) -> Vec<i32> {
        let mut items = items;
        let mut answer = vec![0; queries.len()];

        items.sort_unstable_by_key(|item| (item[0], -item[1]));

        for i in 1..items.len() {
            items[i][1] = items[i][1].max(items[i - 1][1]);
        }

        for i in 0..queries.len() {
            let j = items
                .binary_search(&vec![queries[i], i32::MAX])
                .unwrap_err();

            answer[i] = match j {
                0 => 0,
                _ => items[j - 1][1],
            };
        }

        answer
    }
}