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Leetcode #2389: Longest Subsequence With Limited Sum

In this guide, we solve Leetcode #2389 Longest Subsequence With Limited Sum in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

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].

Quick Facts

  • Difficulty: Easy
  • Premium: No
  • Tags: Greedy, Array, Binary Search, Prefix Sum, Sorting

Intuition

The problem structure suggests a monotonic decision, which makes binary search a natural fit.

By halving the search space each step, we reach the answer efficiently.

Approach

Search either directly on a sorted array or on the answer space using a check function.

Each check is fast, and the logarithmic search keeps the overall runtime low.

Steps:

  • Define the search bounds.
  • Check the mid point condition.
  • Narrow the bounds until convergence.

Example

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.

Python Solution

class Solution: def answerQueries(self, nums: List[int], queries: List[int]) -> List[int]: nums.sort() s = list(accumulate(nums)) return [bisect_right(s, q) for q in queries]

Complexity

The time complexity is O((n+m)×log⁡n)O((n + m) \times \log n)O((n+m)×logn), and the space complexity is O(n)O(n)O(n) or O(log⁡n)O(\log n)O(logn). The space complexity is O(n)O(n)O(n) or O(log⁡n)O(\log n)O(logn).

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.


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