Formatted question description: https://leetcode.ca/all/2070.html

# 2070. Most Beautiful Item for Each Query (Medium)

You are given a 2D integer array `items`

where `items[i] = [price`

denotes the _{i}, beauty_{i}]**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 *`j`

^{th}* 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 <= price`

_{i}, beauty_{i}, queries[j] <= 10^{9}

**Similar Questions**:

## Solution 1. Offline Query

Sort both `items`

and `queries`

in ascending order of `price`

. Traverse `queries`

. Use a pointer `i`

to cover all the items with a price `<=`

the current query, and compute the maximum beauty among them. In this way, we traverse both array only once.

```
// OJ: https://leetcode.com/problems/most-beautiful-item-for-each-query/
// 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;
}
};
```

## Discuss

https://leetcode.com/problems/most-beautiful-item-for-each-query/discuss/1576065/C%2B%2B-Offline-Query