How do you solve Most Beautiful Item for Each Query on LeetCode?

Most Beautiful Item for Each Query (LeetCode #2070) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n + m log n) time complexity and O(n) space complexity. Key insight: Sorting the items allows us to efficiently find the maximum beauty for each price threshold.

What is the time complexity of Most Beautiful Item for Each Query?

The optimal solution for Most Beautiful Item for Each Query runs in O(n log n + m log n) time and uses O(n) space. Sorting the items takes O(n log n), and each query takes O(log n) due to binary search, resulting in O(m log n) for m queries.

What is the brute force solution for Most Beautiful Item for Each Query?

The brute force approach for Most Beautiful Item for Each Query is Brute Force, with O(n²) time complexity and O(1) space complexity. For each query, we will check all items to find the maximum beauty of items that have a price less than or equal to the query price. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n log n + m log n).

What algorithm or data structure is used to solve Most Beautiful Item for Each Query?

Most Beautiful Item for Each Query uses the following concepts: Array, Binary Search, Sorting. The recommended patterns to study are: Sorting, Binary Search, Prefix Maximum.

What are the interview tips for Most Beautiful Item for Each Query?

Always consider sorting as a first step when dealing with ranges or thresholds. Think about how you can preprocess data to make query responses faster. Practice binary search techniques, as they are commonly used in optimization problems.

What are common mistakes when solving Most Beautiful Item for Each Query?

Not sorting the items before processing queries, leading to incorrect results. Failing to handle cases where no items are found for a query, which should return 0.

#2070

Most Beautiful Item for Each Query

Medium

ArrayBinary SearchSortingSortingBinary SearchPrefix Maximum

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n + m log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n + m log n)Space O(n)

By sorting the items based on price and using a prefix maximum array for beauty, we can efficiently answer each query in logarithmic time using binary search.

⚙️

Algorithm

4 steps

1Step 1: Sort the items by price.
2Step 2: Create a prefix maximum array for beauty values, where each index corresponds to the maximum beauty up to that price.
3Step 3: For each query, use binary search to find the rightmost item with a price less than or equal to the query price.
4Step 4: Return the maximum beauty from the prefix array for the found index; if no valid index is found, return 0.

solution.py11 lines

1def maxBeauty(items, queries):
2    items.sort()
3    max_beauty = [0] * len(items)
4    max_beauty[0] = items[0][1]
5    for i in range(1, len(items)):
6        max_beauty[i] = max(max_beauty[i-1], items[i][1])
7    result = []
8    for q in queries:
9        idx = bisect.bisect_right(items, [q, float('inf')]) - 1
10        result.append(max_beauty[idx] if idx >= 0 else 0)
11    return result

ℹ

Complexity note: Sorting the items takes O(n log n), and each query takes O(log n) due to binary search, resulting in O(m log n) for m queries.

1Sorting the items allows us to efficiently find the maximum beauty for each price threshold.
2Using a prefix maximum array helps in quickly retrieving the maximum beauty without re-evaluating items.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.