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Maximum Sum With Exactly K Elements

Number: 2767

Difficulty: Easy

Paid? No

Companies: N/A

Problem Description

Given a 0-indexed integer array nums and an integer k, perform exactly k operations to maximize your score. In each operation, select an element m from nums, remove it, add a new element with a value of m+1 to the array, and increase your score by m. Return the maximum score achievable after performing the operation k times.

Key Insights

  • Greedy approach works because selecting the maximum element at each step yields the highest immediate gain.
  • The operation turns the selected maximum element m into m+1 which will be higher than the current maximum in subsequent operations.
  • Instead of simulating each operation, you can directly calculate the sum of picking the maximum element and then its subsequent increments.
  • The arithmetic progression of chosen elements is: max, max+1, ..., max+k-1.

Space and Time Complexity

Time Complexity: O(n) to find the maximum element in the nums array.
Space Complexity: O(1)

Solution

The optimal strategy is to always choose the maximum element available, which upon selection becomes m+1. Over k operations, if max is the initial maximum element, the chosen elements will be max, max+1, ..., max+k-1. The total maximum score becomes:

score = k * max + (k*(k-1))/2

This uses the formula for the sum of an arithmetic progression. The solution requires a single pass through the array to find the maximum value (O(n) time) and then constant time to calculate the answer.

Code Solutions

# Function to compute the maximum score
def maximize_score(nums, k):
    # Find the maximum element in the array
    max_val = max(nums)
    # Calculate the score using the arithmetic progression sum formula:
    # max, max+1, ..., max+k-1 -> Total = k * max + (k * (k - 1)) // 2
    return k * max_val + (k * (k - 1)) // 2

# Example usage:
if __name__ == "__main__":
    nums = [1, 2, 3, 4, 5]
    k = 3
    print(maximize_score(nums, k))  # Expected output: 18
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