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2070. Most Beautiful Item for Each Query

Description

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

Solutions

  • class Solution {
        public int[] maximumBeauty(int[][] items, int[] queries) {
            Arrays.sort(items, (a, b) -> a[0] - b[0]);
            for (int i = 1; i < items.length; ++i) {
                items[i][1] = Math.max(items[i - 1][1], items[i][1]);
            }
            int n = queries.length;
            int[] ans = new int[n];
            for (int i = 0; i < n; ++i) {
                int left = 0, right = items.length;
                while (left < right) {
                    int mid = (left + right) >> 1;
                    if (items[mid][0] > queries[i]) {
                        right = mid;
                    } else {
                        left = mid + 1;
                    }
                }
                if (left > 0) {
                    ans[i] = items[left - 1][1];
                }
            }
            return ans;
        }
    }
    
  • class Solution {
    public:
        vector<int> maximumBeauty(vector<vector<int>>& items, vector<int>& queries) {
            sort(items.begin(), items.end());
            for (int i = 1; i < items.size(); ++i) items[i][1] = max(items[i - 1][1], items[i][1]);
            int n = queries.size();
            vector<int> ans(n);
            for (int i = 0; i < n; ++i) {
                int left = 0, right = items.size();
                while (left < right) {
                    int mid = (left + right) >> 1;
                    if (items[mid][0] > queries[i])
                        right = mid;
                    else
                        left = mid + 1;
                }
                if (left) ans[i] = items[left - 1][1];
            }
            return ans;
        }
    };
    
  • 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
    
    
  • func maximumBeauty(items [][]int, queries []int) []int {
    	sort.Slice(items, func(i, j int) bool {
    		return items[i][0] < items[j][0]
    	})
    	for i := 1; i < len(items); i++ {
    		items[i][1] = max(items[i-1][1], items[i][1])
    	}
    	n := len(queries)
    	ans := make([]int, n)
    	for i, v := range queries {
    		left, right := 0, len(items)
    		for left < right {
    			mid := (left + right) >> 1
    			if items[mid][0] > v {
    				right = mid
    			} else {
    				left = mid + 1
    			}
    		}
    		if left > 0 {
    			ans[i] = items[left-1][1]
    		}
    	}
    	return ans
    }
    

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