2070. Most Beautiful Item for Each Query

2070. Most Beautiful Item for Each Query

Difficulty: Medium

Topics: Array, Binary Search, Sorting

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

Hint:

1. Can we process the queries in a smart order to avoid repeatedly checking the same items?


2. How can we use the answer to a query for other queries?

Solution:

We can use sorting and binary search techniques. Here’s the plan:

Approach

1. 
   

   Sort the Items by Price:

   
   
   
   
   
   • First, sort items by their price. This way, as we iterate through the items, we can keep track of the maximum beauty seen so far for items up to any given price.
   
   
   
   


2. 
   

   Sort the Queries with their Original Indices:

   
   
   
   
   
   • Create an array of queries paired with their original indices, then sort this array by the query values.
   
   
   • Sorting helps because we can process queries in increasing order of price and avoid recalculating beauty values for lower prices repeatedly.
   
   
   
   


3. 
   

   Iterate through Items and Queries Simultaneously:

   
   
   
   
   
   • Using two pointers, process each query:
     
     
     
     
     • For each query, move the pointer through items with a price less than or equal to the query’s price.
     
     
     • Track the maximum beauty as you go through these items, and use this value to answer the current query.
     
     
     • This avoids repeatedly checking items for multiple queries.
     
     
     
     
   
   
   
   


4. 
   

   Store and Return Results:

   
   
   
   
   
   • Once processed, store the maximum beauty result for each query based on the original index to maintain the order.
   
   
   • Return the answer array.
   
   
   
   

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