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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_{i}, beauty_{i}]
denotes the price and beauty of an item respectively.
You are also given a 0indexed 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/mostbeautifulitemforeachquery/ // 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; } };

// Todo

class Solution: def maximumBeauty(self, items: List[List[int]], queries: List[int]) > List[int]: items.sort() prices = [p for p, _ in items] mx = [items[0][1]] for _, b in items[1:]: mx.append(max(mx[1], b)) ans = [0] * len(queries) for i, q in enumerate(queries): j = bisect_right(prices, q) if j: ans[i] = mx[j  1] return ans ############ # 2070. Most Beautiful Item for Each Query # https://leetcode.com/problems/mostbeautifulitemforeachquery/ class Solution: def maximumBeauty(self, items: List[List[int]], queries: List[int]) > List[int]: n = len(items) items.sort(key = lambda x : (x[0], x[1])) res = [] A = [(0, 0)] curr = 0 for i, (price, beauty) in enumerate(items): if i > 1 and price == items[i  1][0]: continue curr = max(curr, beauty) A.append((price, curr)) for q in queries: index = bisect.bisect(A, (q + 1, ))  1 res.append(A[index][1]) return res
Discuss
https://leetcode.com/problems/mostbeautifulitemforeachquery/discuss/1576065/C%2B%2BOfflineQuery