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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

  • 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*(k-1)/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
        }
    }
    

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