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

C++

// Time: O(NlogN + MlogM + M + N) where `N`/`M` is the length of `items`/`queries`.
// Space: O(M)
class Solution {
public:
    vector<int> maximumBeauty(vector<vector<int>>& A, vector<int>& Q) {
        vector<pair<int, int>> queries;
        for (int i = 0; i < Q.size(); ++i) queries.push_back({Q[i],i});
        sort(begin(queries), end(queries));
        sort(begin(A), end(A));
        int i = 0, N = A.size(), maxBeauty = 0;
        vector<int> ans(Q.size());
        for (auto q : queries) {
            auto &[query, index] = q;
            for (; i < N && A[i][0] <= query; ++i) maxBeauty = max(maxBeauty, A[i][1]);
            ans[index] = maxBeauty;
        }
        return ans;
    }
};

JAVA

• class Solution {
      public int[] maximumBeauty(int[][] items, int[] queries) {
          Arrays.sort(items, (a, b) -> a[0] - b[0]);
          for (int i = 1; i < items.length; ++i) {
              items[i][1] = Math.max(items[i - 1][1], items[i][1]);
          }
          int n = queries.length;
          int[] ans = new int[n];
          for (int i = 0; i < n; ++i) {
              int left = 0, right = items.length;
              while (left < right) {
                  int mid = (left + right) >> 1;
                  if (items[mid][0] > queries[i]) {
                      right = mid;
                  } else {
                      left = mid + 1;
                  }
              }
              if (left > 0) {
                  ans[i] = items[left - 1][1];
              }
          }
          return ans;
      }
  }

PYTHON

from typing import List

class Solution:
    def maximumBeauty(self, items: List[List[int]], queries: List[int]) -> List[int]:
        queries_with_indices = [(queries[i], i) for i in range(len(queries))]
        queries_with_indices.sort()
        items.sort()
        
        i, n = 0, len(items)
        max_beauty = 0
        ans = [0] * len(queries)
        
        for query, index in queries_with_indices:
            while i < n and items[i][0] <= query:
                max_beauty = max(max_beauty, items[i][1])
                i += 1
            ans[index] = max_beauty
            
        return ans