2070. Most Beautiful Item for Each Query

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 <= 10^5

• items[i].length == 2

• 1 <= pricei, beautyi, queries[j] <= 10^9

Solution (Java)

class Solution {
    public int[] maximumBeauty(int[][] items, int[] queries) {
        int[] res = new int[queries.length];
        Arrays.sort(items, Comparator.comparingInt(a -> a[1]));
        for (int i = 0; i < res.length; i++) {
            res[i] = maxBeauty(items, queries[i]);
        }
        return res;
    }

    private int maxBeauty(int[][] items, int query) {
        for (int i = items.length - 1; i >= 0; i--) {
            int price = items[i][0];
            int beauty = items[i][1];
            if (price <= query) {
                return beauty;
            }
        }
        return 0;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).