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

Solution: Prefix Max + Binary Search

Sort items by price. For each price, use a treemap to store the max beauty of an item whose prices is <= p. Then use binary search to find the max beauty whose price is <= p.

Time complexity: Pre-processing O(nlogn) + query: O(qlogn)
Space complexity: O(n)

// Author: Huahua
class Solution {
public:
  vector<int> maximumBeauty(vector<vector<int>>& items, vector<int>& queries) {
    sort(begin(items), end(items));
    map<int, int> m;
    int best = 0;
    for (const auto& item : items) {
      best = max(best, item[1]);
      m[item[0]] = best;
    }
    vector<int> ans;
    for (const int q: queries) {
      auto it = m.upper_bound(q);
      if (it == begin(m))
        ans.push_back(0);
      else
        ans.push_back(prev(it)->second);
    }
    return ans;
  }
};

The post 花花酱 LeetCode 2070. Most Beautiful Item for Each Query appeared first on Huahua's Tech Road.