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Leetcode #1403: Minimum Subsequence in Non-Increasing Order

In this guide, we solve Leetcode #1403 Minimum Subsequence in Non-Increasing Order 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

Given the array nums, obtain a subsequence of the array whose sum of elements is strictly greater than the sum of the non included elements in such subsequence. If there are multiple solutions, return the subsequence with minimum size and if there still exist multiple solutions, return the subsequence with the maximum total sum of all its elements.

Quick Facts

  • Difficulty: Easy
  • Premium: No
  • Tags: Greedy, Array, Sorting

Intuition

A locally optimal choice leads to a globally optimal result for this structure.

That means we can commit to decisions as we scan without backtracking.

Approach

Sort or preprocess if needed, then repeatedly take the best available local choice.

Maintain the minimal state necessary to validate the greedy decision.

Steps:

  • Sort or preprocess as needed.
  • Iterate and pick the best local option.
  • Track the current solution.

Example

Input: nums = [4,3,10,9,8] Output: [10,9] Explanation: The subsequences [10,9] and [10,8] are minimal such that the sum of their elements is strictly greater than the sum of elements not included. However, the subsequence [10,9] has the maximum total sum of its elements.

Python Solution

class Solution: def minSubsequence(self, nums: List[int]) -> List[int]: ans = [] s, t = sum(nums), 0 for x in sorted(nums, reverse=True): t += x ans.append(x) if t > s - t: break return ans

Complexity

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