Maximum Sum With Exactly K Elements

Given an array nums, we have to repeat an operation k times where we:

  • Select the largest element m from nums.
  • Remove the selected element m.
  • Add a new element with value m + 1.
  • Increase our score by m.

After repeating this operation k times, we need to return the maximum score. To maximize the score, we should always select the largest element from the array in each step.

Steps

  1. Initialize Score: Initialize the score as 0.
  2. Repeat Operation k Times: Repeat the following steps k times: a. Find the maximum element m in the array. b. Add m to the score. c. Remove m from the array and add m + 1 to the array.
  3. Return the Score: Return the final score.

Code

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class Solution:
    def maximizeSum(self, nums: List[int], k: int) -> int:
        # Step 1: Initialize the score
        score = 0

        # Step 2: Repeat operation k times
        for _ in range(k):
            m = max(nums)       # Find the maximum element
            score += m          # Add m to the score
            nums.remove(m)      # Remove m from the array
            nums.append(m + 1)  # Add m + 1 to the array

        # Step 3: Return the score
        return score

Example

For nums = [1,2,3,4,5], k = 3, the steps would be:

  • Iteration 1: Choose 5, Score is 5, nums = [1,2,3,4,6]
  • Iteration 2: Choose 6, Score is 11, nums = [1,2,3,4,7]
  • Iteration 3: Choose 7, Score is 18, nums = [1,2,3,4,8]

So the result is 18.

Complexity

The solution’s time complexity is O(k * n), where n is the length of the array, and space complexity is O(1) as we are not using any additional space other than input. It respects the given constraints, so it will work for the provided input limits.