Maximum Sum With Exactly K Elements
A 0-indexed integer array nums and
an integer k are given. The task is to perform the following
operation exactly k times
in order to maximize your score:
- Select an element m from nums.
- 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.
The maximum score that can
be achieved after performing the operation exactly k times must be returned.
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
class Solution {
public int maximizeSum(int[] nums, int k) {
if (nums == null || nums.length == 0 || k <= 0) return 0;
Arrays.sort(nums);
int sum = 0;
int val = nums[nums.length-1];
for (int i = 0; i < k; i++){
sum += val;
val += 1;
}
return sum;
}
}
Nums = [3], k = 3 => sum = 12
Nums = [1,2,3], k = 3 => sum = 12
Nums = [-1,-1,-1], k = 3 => sum = 0
Nums = [-1,0,1], k = 1 => sum = 1
