LeetCode每日一题(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

---

items 中只有 beauty 与 price 成正比的 item 才是有效的, 我们以 price 和 beauty 对 items 进行排序， 然后遍历 items, 当 items[i].price > items[i-1].price 时， 如果 items[i].beauty < items[i-1].beauty 则我们认为 items[i]不是合法的 item, 需要过滤掉， 当 items[i].price == items[i-1].price 且 items[i].beauty > items[i-1].beauty 则我们认为 items[i-1]没有存在的必要， 需要过滤掉。我们将剩余的 items 放到 BinaryHeap 中， 这样， 在栈顶的永远都是 price 最大且 beauty 最大的 item. 然后我们再将 queries 进行排序， 使我们按照 price 从大到小的顺序进行查询. 这样如果当前 item.price>query.price 则我们需要将 item 弹出栈， 因为后面的 query.price 只会越来越小。如果当前的 item.price<=query.price, 我们则认为 item.beauty 就是最大的 beauty, 直接加到答案中即可。

---

use std::collections::BinaryHeap;

impl Solution {
    pub fn maximum_beauty(mut items: Vec<Vec<i32>>, queries: Vec<i32>) -> Vec<i32> {
        items.sort_by_key(|l| (l[0], l[1]));
        let mut item_heap = BinaryHeap::new();
        let mut prev = items.remove(0);
        for item in items {
            if item[0] > prev[0] {
                if item[1] > prev[1] {
                    item_heap.push((prev[0], prev[1]));
                    prev = item;
                }
                continue;
            }
            if item[1] > prev[1] {
                prev = item;
            }
        }
        item_heap.push((prev[0], prev[1]));
        let mut ans = vec![0; queries.len()];
        let mut query_heap: BinaryHeap<(i32, usize)> = queries
            .into_iter()
            .enumerate()
            .map(|(i, v)| (v, i))
            .collect();
        'outer: while let Some((limit, i)) = query_heap.pop() {
            while let Some((price, beauty)) = item_heap.pop() {
                if price > limit {
                    continue;
                }
                ans[i] = beauty;
                item_heap.push((price, beauty));
                continue 'outer;
            }
            break;
        }
        ans
    }
}