Problem
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:
- Select an element 
mfromnums. - Remove the selected element 
mfrom the array. - Add a new element with a value of 
m + 1to 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 <= 1001 <= nums[i] <= 1001 <= k <= 100
.spoilerbutton {display:block; border:dashed; padding: 0px 0px; margin:10px 0px; font-size:150%; font-weight: bold; color:#000000; background-color:cyan; outline:0; } .spoiler {overflow:hidden;} .spoiler > div {-webkit-transition: all 0s ease;-moz-transition: margin 0s ease;-o-transition: all 0s ease;transition: margin 0s ease;} .spoilerbutton[value="Show Message"] + .spoiler > div {margin-top:-500%;} .spoilerbutton[value="Hide Message"] + .spoiler {padding:5px;}
Solution (Java)
class Solution {
    public int maximizeSum(int[] nums, int k) {
        int max = 0;
        for(int i = 0; i<nums.length; i++){
            max = Math.max(max, nums[i]);
        }
        int ans = 0;
        while(k>0){
            ans+=max;
            max = max+1;
            k--;
        }
        return ans;
    }
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
 - Space complexity : O(n).