# 2656. Maximum Sum With Exactly K Elements

## Description

You are given a 0-indexed integer array nums and an integer k. Your task is to perform the following operation exactly k times in order to maximize your score:

1. Select an element m from nums.
2. Remove the selected element m from the array.
3. Add a new element with a value of m + 1 to the array.
4. Increase your score by m.

Return the maximum score you can achieve after performing the operation exactly k times.

Example 1:

Input: nums = [1,2,3,4,5], k = 3
Output: 18
Explanation: 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
Explanation: 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

## Solutions

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, $k \times x + (k - 1) \times k / 2$.

Time complexity is $O(n)$, where $n$ is the length of the array. Space complexity is $O(1)$.

• class Solution {
public int maximizeSum(int[] nums, int k) {
int x = 0;
for (int v : nums) {
x = Math.max(x, v);
}
return k * x + k * (k - 1) / 2;
}
}

• class Solution {
public:
int maximizeSum(vector<int>& nums, int k) {
int x = *max_element(nums.begin(), nums.end());
return k * x + k * (k - 1) / 2;
}
};

• class Solution:
def maximizeSum(self, nums: List[int], k: int) -> int:
x = max(nums)
return k * x + k * (k - 1) // 2


• func maximizeSum(nums []int, k int) int {
x := slices.Max(nums)
return k*x + k*(k-1)/2
}

• function maximizeSum(nums: number[], k: number): number {
const x = Math.max(...nums);
return k * x + (k * (k - 1)) / 2;
}


• impl Solution {
pub fn maximize_sum(nums: Vec<i32>, k: i32) -> i32 {
let mut mx = 0;

for &n in &nums {
if n > mx {
mx = n;
}
}

((0 + k - 1) * k) / 2 + k * mx
}
}