LeetCode 2656. Maximum Sum With Exactly K Elements Solution in Java, C++, Python & More | Explanation + Code

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

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 kth time we select x+k-1. 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+…+(k-1), that is, k × x + (k - 1) × k / 2.

Time complexity is O(n), where n is the length of the array. Space complexity is O(1).

PythonJavaC++GoTypeScriptRust

class Solution: def maximizeSum(self, nums: List[int], k: int) -> int: x = max(nums) return k * x + k * (k - 1) // 2(code-box)

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; } }(code-box)

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; } };(code-box)

func maximizeSum(nums []int, k int) int { x := slices.Max(nums) return k*x + k*(k-1)/2 }(code-box)

function maximizeSum(nums: number[], k: number): number { const x = Math.max(...nums); return k * x + (k * (k - 1)) / 2; }(code-box)

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 } }(code-box)

Solution 2

Rust

impl Solution { pub fn maximize_sum(nums: Vec<i32>, k: i32) -> i32 { let mx = *nums.iter().max().unwrap_or(&0); ((0 + k - 1) * k) / 2 + k * mx } }(code-box)