Today I have done leetcode's November LeetCoding Challenge with cpp.

November LeetCoding Challenge 12

Description

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

Solution

class Solution {
  using vi = vector<int>;
public:
  vector<int> maximumBeauty(vector<vector<int>>& items, vector<int>& queries) {
    sort(items.begin(), items.end(), [](const vi &a, const vi &b) {
      return a[0] < b[0] || (a[0] == b[0] && a[1] > b[1]);
    });
    int prevMax = INT_MIN;
    for(auto &item : items) {
      item[1] = max(prevMax, item[1]);
      prevMax = max(item[1], prevMax);
    }
    items.resize(unique(items.begin(), items.end()) - items.begin());
    vector<int> answer;
    answer.reserve(queries.size());
    for(auto q : queries) {
      auto it = upper_bound(items.begin(), items.end(), vector<int>({q, INT_MAX}));
      if(it == items.begin()) {
        answer.push_back(0);
      } else {
        --it;
        answer.push_back((*it)[1]);
      }
    }
    return answer;
  }
};