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2656. Maximum Sum With Exactly K Elements
Description
You are given a 0indexed integer array nums
and an integer k
. Your task is to perform the following operation exactly k
times in order to maximize your score:
 Select an element
m
fromnums
.  Remove the selected element
m
from the array.  Add a new element with a value of
m + 1
to the array.  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 $x+k1$. 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 $0+1+2+\cdots+(k1)$, 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 := 0 for _, v := range nums { x = max(x, v) } return k*x + k*(k1)/2 } func max(a, b int) int { if a > b { return a } return b }

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 } }