Introduction

In this blog post, we will explore how to solve the LeetCode problem "2070" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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

Explanation

Here's the solution to the problem:

• Sort and Precompute Maximums: Sort the items based on price. Then, create a new array where each element represents the maximum beauty seen so far up to that index. This precomputation enables efficient binary search later.
• Binary Search for Each Query: For each query, use binary search on the sorted prices to find the index of the last item whose price is less than or equal to the query price. The corresponding beauty value in the precomputed maximum beauty array will be the answer.

• Handle No Item Found: If the binary search returns -1 (no item satisfies the condition), the answer is 0.

• Runtime Complexity: O(n log n + m log n), where n is the number of items and m is the number of queries. Sorting takes O(n log n), precomputing maximums takes O(n), and performing binary search for each query takes O(m log n).

• Storage Complexity: O(n), due to the sorted items array and the precomputed maximum beauty array.

Code

    def maximumBeauty(items, queries):
    """
    Finds the maximum beauty of an item whose price is less than or equal to each query.

    Args:
        items: A 2D integer array where items[i] = [pricei, beautyi].
        queries: A 0-indexed integer array of queries.

    Returns:
        An array answer of the same length as queries where answer[j] is the answer to the jth query.
    """

    items.sort()  # Sort items based on price
    max_beauties = []
    current_max = 0
    for price, beauty in items:
        current_max = max(current_max, beauty)
        max_beauties.append(current_max)

    answer = []
    for query in queries:
        # Binary search to find the index of the last item with price <= query
        left, right = 0, len(items) - 1
        index = -1
        while left <= right:
            mid = (left + right) // 2
            if items[mid][0] <= query:
                index = mid
                left = mid + 1
            else:
                right = mid - 1

        if index == -1:
            answer.append(0)
        else:
            answer.append(max_beauties[index])

    return answer