# Solving Leetcode Interviews in Seconds with AI: Longest Subsequence With Limited Sum


	# 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](https://www.chatmagic.app), 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
	```python
	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
	```