algoadvance

You are given an integer array nums and an integer k. Return the subsequence of nums that has the largest sum and is of length k.

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.

Clarifying Questions

  1. Are there any constraints on the values in nums?
    • The element values in nums can be positive, negative or zero.
  2. Can k be greater than the length of nums?
    • No, k will always be less than or equal to the length of nums.
  3. Can nums contain duplicate values?
    • Yes, nums can contain duplicate values.
  4. What should be returned if nums is empty?
    • The problem states that k <= len(nums), so this scenario shouldn’t occur.

Strategy

  1. Identify the k largest elements:
    • We need to identify the k largest numbers in the array. This can be efficiently done using a min-heap of size k. Python’s heapq can be helpful here.
  2. Maintain the original order:
    • Once we have identified the k largest elements, we need to extract them in the original order they appear in the array. This means we need to keep track of the indices of these elements.
  3. Algorithm:
    • Create a list of tuples containing each element and its index.
    • Sort the list of tuples primarily by value in descending order, but essentially we’ll construct a max-heap to easily get the top k elements.
    • From these tuples, pick the top k elements.
    • Sort the picked elements by their original indices to maintain the order they appear in the original array.
    • Extract the values from these sorted tuples.

Code

import heapq

def maxSubsequence(nums, k):
    # Create a list of tuples (value, index).
    enumerated_nums = [(num, i) for i, num in enumerate(nums)]

    # Get the k largest elements along with their indices using a heap.
    largest_k = heapq.nlargest(k, enumerated_nums, key=lambda x: x[0])

    # Sort these k elements by their original indices to maintain the order.
    largest_k.sort(key=lambda x: x[1])

    # Extract the values from the sorted tuples.
    return [num for num, _ in largest_k]

# Example usage:
nums = [3, 4, 3, 3]
k = 2
print(maxSubsequence(nums, k))  # Output: [4, 3]

Time Complexity

  1. Enumeration and Tuple Creation:
    • This step takes O(n), where n is the number of elements in nums.
  2. Finding k Largest Elements:
    • Using heapq.nlargest, this step takes O(n log k).
  3. Sorting k Elements by Index:
    • This step takes O(k log k).
  4. Final Extraction:
    • This step takes O(k).

The overall time complexity is dominated by the heapq.nlargest operation, i.e., O(n log k).

By following this approach, we ensure that we are efficiently finding the required subsequence while maintaining the original order of elements.

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