2656

2656. Maximum Sum With Exactly K Elements (Easy)

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

 

Related Topics:
Array, Greedy

Solution 1. Greedy

Greedily pick the greatest element n, and keep increasing it. We will get n + (n+1) + ... + (n+k-1) = (n + n + k - 1) * k / 2 score.

// OJ: https://leetcode.com/problems/maximum-sum-with-exactly-k-elements
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int maximizeSum(vector<int>& A, int k) {
        int n = *max_element(begin(A), end(A));
        return (n + n + k - 1) * k / 2;
    }
};