# Leetcode – Minimum Subsequence in Non-Increasing Order Solution

One of the many fascinating problems on Leetcode is the “Minimum Subsequence in Non-Increasing Order”. This problem allows us to explore multiple facets of algorithm design, Python programming, and mathematical logic. In this extensive article, we will dissect the problem in detail, examine various approaches to solve it, and evaluate the pros and cons of each.

1. Understanding the Problem Statement
2. Initial Thoughts and Simplification
3. Brute-Force Approach
4. Greedy Algorithm
5. Built-in Sorting and Slicing
6. Time and Space Complexity Analysis
7. Testing the Solution
8. Further Optimizations and Variants
9. Conclusion

## 1. Understanding the Problem Statement

The problem asks for the smallest subsequence of a given array such that the sum of the subsequence is strictly greater than the sum of the numbers not included in the subsequence. Additionally, the elements of the subsequence must be sorted in non-increasing order.

### Constraints and Assumptions

• The array contains n integers where 1 <= n <= 500.
• Each integer nums[i] is 1 <= nums[i] <= 100.

## 2. Initial Thoughts and Simplification

This problem involves sorting, selecting elements, and summing them. The aim is to get the smallest subsequence such that its sum is greater than the sum of the remaining elements.

## 3. Brute-Force Approach

A brute-force solution might involve checking all possible subsequences of the array and determining if their sum is greater than the sum of the remaining elements. However, this would be highly inefficient with a time complexity of O(2^n).

## 4. Greedy Algorithm

A more efficient approach is to use a greedy algorithm. Sort the array in non-increasing order and then keep adding elements to the subsequence until the sum of the subsequence becomes greater than the sum of the remaining elements.

Python Code:

def minSubsequence(nums):
nums.sort(reverse=True)
total_sum = sum(nums)
sub_sum = 0
result = []

for num in nums:
sub_sum += num
result.append(num)
if sub_sum > total_sum - sub_sum:
return result

## 5. Built-in Sorting and Slicing

Python has built-in sorting that works in O(n log n) time. We can take advantage of this, along with list slicing for a more Pythonic solution.

Python Code:

def minSubsequence(nums):
nums.sort()
total_sum = sum(nums)
sub_sum = 0
result = []

while sub_sum <= total_sum:
num = nums.pop()
sub_sum += num
total_sum -= num
result.append(num)

return result

## 6. Time and Space Complexity Analysis

• Brute-Force Approach: Time Complexity is O(2^n) and Space Complexity is O(n).
• Greedy Algorithm: Time Complexity is O(n log n) and Space Complexity is O(n).
• Built-in Sorting and Slicing: Time Complexity is O(n log n) and Space Complexity is O(n).

## 7. Testing the Solution

To ensure that your solution is correct, test with:

• Arrays with minimum and maximum constraints.
• Arrays with all identical elements.
• Arrays with unique elements.
• Random arrays.

## 8. Further Optimizations and Variants

While the greedy algorithm is efficient enough for this problem, one might consider:

• Using quickselect to find the kth largest element and then deriving the subsequence based on that.

## 9. Conclusion

The “Minimum Subsequence in Non-Increasing Order” problem offers an insightful way to understand greedy algorithms and sorting techniques. Multiple approaches, from brute-force to greedy algorithms, have their unique advantages and disadvantages, which are critical to understand for solving similar problems effectively.