Introduction

In this blog post, we will explore how to solve the LeetCode problem "2389" 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 an integer array nums of length n, and an integer array queries of length m. Return an array answer of length m where answer[i] is the maximum size of a subsequence that you can take from nums such that the sum of its elements is less than or equal to queries[i]. A subsequence is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements. Example 1: Input: nums = [4,5,2,1], queries = [3,10,21] Output: [2,3,4] Explanation: We answer the queries as follows: - The subsequence [2,1] has a sum less than or equal to 3. It can be proven that 2 is the maximum size of such a subsequence, so answer[0] = 2. - The subsequence [4,5,1] has a sum less than or equal to 10. It can be proven that 3 is the maximum size of such a subsequence, so answer[1] = 3. - The subsequence [4,5,2,1] has a sum less than or equal to 21. It can be proven that 4 is the maximum size of such a subsequence, so answer[2] = 4. Example 2: Input: nums = [2,3,4,5], queries = [1] Output: [0] Explanation: The empty subsequence is the only subsequence that has a sum less than or equal to 1, so answer[0] = 0. Constraints: n == nums.length m == queries.length 1 <= n, m <= 1000 1 <= nums[i], queries[i] <= 106

Explanation

Here's the breakdown of the solution:

• Sorting: Sort the nums array in ascending order. This allows us to greedily pick the smallest elements first to maximize the subsequence size.
• Prefix Sum: Calculate the prefix sum of the sorted nums array. This enables efficient calculation of the sum of subsequences.

• Binary Search: For each query, use binary search on the prefix sum array to find the largest index (subsequence size) whose sum is less than or equal to the query value.

• Runtime Complexity: O(n log n + m log n), where n is the length of nums and m is the length of queries.

• Storage Complexity: O(n)

Code

    def answerQueries(nums, queries):
    nums.sort()
    n = len(nums)
    prefix_sum = [0] * (n + 1)
    for i in range(n):
        prefix_sum[i + 1] = prefix_sum[i] + nums[i]

    answer = []
    for query in queries:
        l, r = 0, n
        ans = 0
        while l <= r:
            mid = (l + r) // 2
            if prefix_sum[mid] <= query:
                ans = mid
                l = mid + 1
            else:
                r = mid - 1
        answer.append(ans)
    return answer