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

Solution: Binary Search

• Time complexity: O(nlogn)
• Space complexity: O(n)

 

JavaScript

/**
 * @param {number[][]} items
 * @param {number[]} queries
 * @return {number[]}
 */
const maximumBeauty = function (items, queries) {
  const itemsMap = new Map();

  for (const [price, beauty] of items) {
    const itemBeauty = itemsMap.get(price) ?? 0;

    itemsMap.set(price, Math.max(beauty, itemBeauty));
  }
  const n = itemsMap.size;
  const filteredItems = [];

  for (const [price, beauty] of itemsMap) {
    filteredItems.push({ price, beauty });
  }
  filteredItems.sort((a, b) => a.price - b.price);

  for (let index = 1; index < n; index++) {
    const { beauty } = filteredItems[index];
    const maxBeauty = filteredItems[index - 1].beauty;

    filteredItems[index].beauty = Math.max(beauty, maxBeauty);
  }

  const findMaximumBeauty = price => {
    let left = 0;
    let right = n - 1;

    while (left < right) {
      const mid = Math.floor((left + right) / 2);
      const item = filteredItems[mid];

      item.price >= price ? (right = mid) : (left = mid + 1);
    }
    const index = filteredItems[left].price > price ? left - 1 : left;

    return filteredItems[index]?.beauty ?? 0;
  };

  return queries.map(price => findMaximumBeauty(price));
};