2070. 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

class Solution {
public:
vector<int> maximumBeauty(vector<vector<int>>& items,vector<int>& queries) {
vector<int> ans(queries.size());

// Sort and store max beauty
sort(items.begin(), items.end(),
[](vector<int>& a, vector<int>& b) { return a[0] < b[0]; });

int maxBeauty = items[0][1];
for (int i = 0; i < items.size(); i++) {
maxBeauty = max(maxBeauty, items[i][1]);
items[i][1] = maxBeauty;
}

for (int i = 0; i < queries.size(); i++) {
// answer i-th query
ans[i] = binarySearch(items, queries[i]);
}

return ans;
}

int binarySearch(vector<vector<int>>& items, int targetPrice) {
int left = 0;
int right = items.size() - 1;
int maxBeauty = 0;
while (left <= right) {
int mid = (left + right) / 2;
if (items[mid][0] > targetPrice) {
right = mid - 1;
} else {
// Found viable price. Keep moving to right
maxBeauty = max(maxBeauty, items[mid][1]);
left = mid + 1;
}
}
return maxBeauty;
}
};