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 <= 105items[i].length == 21 <= pricei, beautyi, queries[j] <= 109
Similar Questions:
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/
// Author: github.com/lzl124631x
// 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;
}
};