LeetCode Problem Workspace
Maximum Sum With Exactly K Elements
Maximize your score by performing exactly k operations on an array using a greedy approach to select the highest values.
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Practice Focus
Easy · Greedy choice plus invariant validation
Answer-first summary
Maximize your score by performing exactly k operations on an array using a greedy approach to select the highest values.
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In this problem, you are asked to maximize your score by performing exactly k operations on an array. Each operation selects the maximum value available. A greedy approach ensures optimal performance by repeatedly selecting the largest available element.
Problem Statement
You are given a 0-indexed integer array nums and an integer k. To maximize your score, you must perform the following operation exactly k times: select the largest element from nums, add it to your score, and increase that element by 1. After performing the operation k times, return the maximum score you can achieve.
For example, given nums = [1,2,3,4,5] and k = 3, the process involves selecting the highest values three times, which yields a score of 18 after the k operations. The problem requires a greedy approach to solve it efficiently.
Examples
Example 1
Input: nums = [1,2,3,4,5], k = 3
Output: 18
We need to choose exactly 3 elements from nums to maximize the sum. For the first iteration, we choose 5. Then sum is 5 and nums = [1,2,3,4,6] For the second iteration, we choose 6. Then sum is 5 + 6 and nums = [1,2,3,4,7] For the third iteration, we choose 7. Then sum is 5 + 6 + 7 = 18 and nums = [1,2,3,4,8] So, we will return 18. It can be proven, that 18 is the maximum answer that we can achieve.
Example 2
Input: nums = [5,5,5], k = 2
Output: 11
We need to choose exactly 2 elements from nums to maximize the sum. For the first iteration, we choose 5. Then sum is 5 and nums = [5,5,6] For the second iteration, we choose 6. Then sum is 5 + 6 = 11 and nums = [5,5,7] So, we will return 11. It can be proven, that 11 is the maximum answer that we can achieve.
Constraints
- 1 <= nums.length <= 100
- 1 <= nums[i] <= 100
- 1 <= k <= 100
Solution Approach
Greedy Approach
The key to solving this problem is using a greedy approach. At each step, choose the largest element from the array to maximize the score. This approach ensures that you are always making the most beneficial choice by selecting the largest available element.
Invariant Validation
Maintain an invariant by ensuring that after each operation, the updated array reflects the new value of the selected element. This invariant allows you to properly track the changes in the array and ensures that subsequent choices are based on accurate data.
Time Complexity Considerations
Using a heap or a sorting method to track the largest element is crucial for optimizing performance. Sorting the array initially or using a priority queue will allow efficient access to the maximum value, ensuring that the solution runs within acceptable time limits for large inputs.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
The time complexity of the solution depends on the method used for selecting the largest element. Sorting the array takes O(n log n) time, while using a priority queue allows for O(k log n) time complexity, where n is the number of elements in the array and k is the number of operations to be performed.
What Interviewers Usually Probe
- Candidate efficiently handles greedy algorithms and invariant tracking.
- Candidate considers performance and time complexity when choosing data structures.
- Candidate demonstrates the ability to optimize code using sorting or priority queues.
Common Pitfalls or Variants
Common pitfalls
- Failing to properly update the array after each operation, leading to incorrect choices.
- Using inefficient data structures, causing the solution to exceed time limits.
- Not validating the greedy approach thoroughly for all edge cases, especially when k is large.
Follow-up variants
- Modify the problem to allow for selecting smaller elements and track the resulting score.
- Consider the problem where the number of operations k is variable and depends on the input.
- Solve the problem by using dynamic programming instead of a greedy approach.
FAQ
What is the greedy approach for the Maximum Sum With Exactly K Elements problem?
The greedy approach involves repeatedly selecting the largest element from the array to maximize the score after performing exactly k operations.
How can I efficiently find the largest element in the array?
You can use a priority queue (heap) or sort the array to efficiently track and select the largest element.
What happens if I don't update the array after each operation?
Failing to update the array after each operation will result in incorrect subsequent selections and lead to an incorrect solution.
Can this problem be solved using dynamic programming?
While the greedy approach is optimal for this problem, dynamic programming could be considered for variants or more complex modifications.
How does GhostInterview help me solve the Maximum Sum With Exactly K Elements problem?
GhostInterview provides structured guidance to understand the problem's greedy nature, select efficient data structures, and optimize performance during interviews.
Solution
Solution 1: Greedy + Mathematics
We notice that to make the final score maximum, we should make each choice as large as possible. Therefore, we select the largest element $x$ in the array for the first time, $x+1$ for the second time, $x+2$ for the third time, and so on, until the $k$th time we select $x+k-1$. This way of selection ensures that the element selected each time is the largest in the current array, so the final score is also the largest. The answer is $k$ $x$ sum plus $0+1+2+\cdots+(k-1)$, that is, $k \times x + (k - 1) \times k / 2$.
class Solution:
def maximizeSum(self, nums: List[int], k: int) -> int:
x = max(nums)
return k * x + k * (k - 1) // 2Solution 2
#### Rust
class Solution:
def maximizeSum(self, nums: List[int], k: int) -> int:
x = max(nums)
return k * x + k * (k - 1) // 2Continue Topic
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