Question

2070. Most Beautiful Item for Each Query

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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

Solution 1

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

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;
  }
};

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
}