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] = [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

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;
}
};

• // 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/most-beautiful-item-for-each-query/

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/most-beautiful-item-for-each-query/discuss/1576065/C%2B%2B-Offline-Query