How do you solve Minimum Subsequence in Non-Increasing Order on LeetCode?

Minimum Subsequence in Non-Increasing Order (LeetCode #1403) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Sorting helps in quickly identifying the largest elements to form a valid subsequence.

What is the time complexity of Minimum Subsequence in Non-Increasing Order?

The optimal solution for Minimum Subsequence in Non-Increasing Order runs in O(n log n) time and uses O(n) space. The time complexity is O(n log n) due to the sorting step, while the space complexity is O(n) for storing the result subsequence.

What is the brute force solution for Minimum Subsequence in Non-Increasing Order?

The brute force approach for Minimum Subsequence in Non-Increasing Order is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves generating all possible subsequences and checking their sums. This is straightforward but inefficient, as it explores every combination of elements. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Minimum Subsequence in Non-Increasing Order?

Minimum Subsequence in Non-Increasing Order uses the following concepts: Array, Greedy, Sorting. The recommended patterns to study are: Greedy, Sorting.

What are the interview tips for Minimum Subsequence in Non-Increasing Order?

Always clarify the requirements of the problem, especially regarding subsequences. Discuss your thought process aloud to demonstrate your understanding of the problem. Consider edge cases, such as arrays with duplicate elements or very small sizes.

What are common mistakes when solving Minimum Subsequence in Non-Increasing Order?

Not considering the condition of strict inequality when summing subsequences. Failing to sort the array before attempting to form the subsequence.

#1403

Minimum Subsequence in Non-Increasing Order

Easy

ArrayGreedySortingGreedySorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

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Intuition

Time O(n log n)Space O(n)

The optimal approach sorts the array in descending order and iteratively adds elements to a subsequence until its sum exceeds the sum of the remaining elements. This is efficient because it directly targets the largest elements first.

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Algorithm

3 steps

1Step 1: Sort the array in non-increasing order.
2Step 2: Initialize two sums: one for the subsequence and one for the remaining elements.
3Step 3: Iterate through the sorted array, adding elements to the subsequence until its sum is greater than the remaining sum.

solution.py16 lines

1# Full working Python code
2def min_subsequence(nums):
3    nums.sort(reverse=True)
4    total_sum = sum(nums)
5    subseq_sum = 0
6    result = []
7
8    for num in nums:
9        subseq_sum += num
10        result.append(num)
11        if subseq_sum > total_sum - subseq_sum:
12            break
13    return result
14
15# Example usage
16print(min_subsequence([4,3,10,9,8]))  # Output: [10, 9]

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting step, while the space complexity is O(n) for storing the result subsequence.

1Sorting helps in quickly identifying the largest elements to form a valid subsequence.
2Iteratively adding elements to the subsequence until the condition is met ensures minimal size.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.