2070. Most Beautiful Item for Each Query

Problem Description

In this problem, we have a list of items, each represented by a pair [price, beauty]. Our goal is to answer a series of queries, each asking for the maximum beauty value among all items whose price is less than or equal to the query value.

If no items fit the criteria for a given query (all items are more expensive than the query value), the answer for that query is 0.

We are asked to return a list of the maximum beauty values corresponding to each query.

Intuition

To solve the problem efficiently, we first notice that we can handle the queries independently. So, we want a quick way to find the maximum beauty for any given price limit.

We approach this by sorting the items by price. Sorting the items allows us to employ a binary search technique to efficiently find the item with the highest beauty below a certain price threshold.

After sorting items, we create two lists: prices, which holds the sorted prices, and mx, which holds the running maximum beauty observed as we iterate through the sorted items. This ensures that for each price prices[i], mx[i] is the maximum beauty of all items with a price less than or equal to prices[i].

The binary search is carried out by using the bisect_right function from Python's bisect module. For each query, bisect_right finds the index j in the sorted prices list such that all prices to the left of j are less than or equal to the query value.

If such an index j is found and is greater than 0, it means there exists at least one item with a price lower than or equal to the query value. We use j - 1 as the index to get the maximum beauty value from the mx list.

Otherwise, if no index j is returned because all items are too expensive, we default the answer for that query to 0.

This algorithm allows us to answer each query in logarithmic time with respect to the number of items, which is desirable when dealing with a large number of items or queries.

Learn more about Binary Search and Sorting patterns.

Solution Approach

The solution uses a mix of sorting, dynamic programming, and binary search to efficiently answer the maximum beauty queries for given price limits.

Here's the step-by-step implementation strategy:

1. Sorting Items: Start by sorting the items based on their price. This is vital because it allows us to leverage binary search later on. Sorting is done using Python's default sorting algorithm, Timsort, which has a time complexity of O(n log n).

2. Extracting Prices and Initializing Maximum Beauty List (mx):

   • Extract the sorted prices into a list called prices.
   • Initialize a list mx, which keeps track of the maximum beauty encountered so far as we iterate through the items. The first element of mx is simply the beauty of the first item in the sorted list.

3. Building a Running Maximum Beauty:

   • Iterate through each item, starting from the second one (since the first element's max beauty is already recorded).
   • For each item, update the mx list with the greater value between the current item's beauty and the last recorded max beauty in mx. This is a form of dynamic programming, where the result of each step is based on the previous step's result.

4. Answering Queries with Binary Search:

   • Initialize an answer list ans of the same size as queries, defaulting all elements to 0.
   • For each query, use the bisect_right function from the bisect module to perform a binary search on prices to find the point where the query value would be inserted while maintaining the list's order.
     • bisect_right returns an index j that is one position past where the query value would be inserted, so a price less than or equal to the query must be at an index before j.
   • If j is not 0, it means an item with a suitable price exists, and the answer for this query is mx[j - 1] - the corresponding max beauty by that price. If j is 0, it means no items are cheaper than the query, and the answer remains 0.

5. Return the Answer List:

   • After all queries have been processed, return the answer list ans filled with the maximum beauties for each respective query.

This approach effectively decouples the item price-beauty relationship from the queries, by pre-computing a list of maximum beauties (mx) that can later be quickly referenced using binary search. This transforms what could be an O(n*m) problem (naively checking n items for each of m queries) into an O(n log n + m log n) problem, where n is the number of items and m the number of queries.

Example Walkthrough

Let's illustrate the solution approach with a small example:

Suppose we have the following items and queries:

• items = [[3, 2], [5, 4], [3, 1], [10, 7]]
• queries = [2, 4, 6]

Following the steps:

1. Sorting Items:

   • We sort items by price: sorted_items = [[3, 2], [3, 1], [5, 4], [10, 7]]

2. Extracting Prices and Initializing Maximum Beauty List (mx):

   • Extract prices: prices = [3, 3, 5, 10]
   • Initialize mx with the maximum beauty of the first item: mx = [2]

3. Building a Running Maximum Beauty:

   • Process the second item: it has the same price but lower beauty, so mx remains the same: mx = [2]
   • Process the third item: new price with higher beauty, update mx: mx = [2, 4]
   • Process the fourth item: new price with higher beauty, update mx: mx = [2, 4, 7]

4. Answering Queries with Binary Search:

   • Initialize an answer list ans with all zeros: ans = [0, 0, 0]
   • Query 1: 2 is less than all prices, therefore ans[0] remains 0.
   • Query 2: 4 is equal to the second price, bisect_right would place it after index 1, so we use mx[0]: ans = [0, 2, 0]
   • Query 3: 6 would fit between indexes 2 and 3, bisect_right returns 3 so we use mx[2]: ans = [0, 2, 4]

5. Return the Answer List:

   • The final answer list reflecting maximum beauties for each query is: ans = [0, 2, 4]

Thus, for the queries [2, 4, 6], the maximum beauty values for items within these price limits are [0, 2, 4], respectively.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the provided code can be broken down into the following parts:

1. Sorting the items list: The items.sort() method is called on the list of items, which typically has a time complexity of O(n * log(n)), where n is the number of items.

2. Creating the prices list: This involves iterating over the sorted items list to build a new list of prices, which will take O(n).

3. Creating the mx list: A single for-loop is used to construct the mx list. This also runs in O(n) time as it iterates over n items once.

4. Answering the queries by binary search: Each query performs a binary search to find the right index in the prices list, which takes O(log(n)). Since this is done for q queries, the total time complexity for this step is O(q * log(n)).

Combining these steps, the overall time complexity is O(n * log(n) + n + n + q * log(n)), which simplifies to O(n * log(n) + q * log(n)) because the linear terms are overshadowed by the n * log(n) term when n is large.

Space Complexity

The space complexity of the code can be analyzed by considering the additional data structures used:

1. The prices list: This consumes O(n) space.

2. The mx list: This also consumes O(n) space.

3. The ans list: Space needed is O(q) for storing the answers for q queries.

Therefore, the overall space complexity is O(n + n + q) which simplifies to O(n + q) as we add the space required for the two lists related to items and the space for the answers to the queries.

Learn more about how to find time and space complexity quickly using problem constraints.