# 2530. Maximal Score After Applying K Operations

## Description

You are given a 0-indexed integer array nums and an integer k. You have a starting score of 0.

In one operation:

1. choose an index i such that 0 <= i < nums.length,
2. increase your score by nums[i], and
3. replace nums[i] with ceil(nums[i] / 3).

Return the maximum possible score you can attain after applying exactly k operations.

The ceiling function ceil(val) is the least integer greater than or equal to val.

Example 1:

Input: nums = [10,10,10,10,10], k = 5
Output: 50
Explanation: Apply the operation to each array element exactly once. The final score is 10 + 10 + 10 + 10 + 10 = 50.


Example 2:

Input: nums = [1,10,3,3,3], k = 3
Output: 17
Explanation: You can do the following operations:
Operation 1: Select i = 1, so nums becomes [1,4,3,3,3]. Your score increases by 10.
Operation 2: Select i = 1, so nums becomes [1,2,3,3,3]. Your score increases by 4.
Operation 3: Select i = 2, so nums becomes [1,1,1,3,3]. Your score increases by 3.
The final score is 10 + 4 + 3 = 17.


Constraints:

• 1 <= nums.length, k <= 105
• 1 <= nums[i] <= 109

## Solutions

Solution 1: Priority Queue (Max Heap)

To maximize the sum of scores, we need to select the element with the maximum value at each step. Therefore, we can use a priority queue (max heap) to maintain the element with the maximum value.

At each step, we take out the element with the maximum value $v$ from the priority queue, add $v$ to the answer, and replace $v$ with $\lceil \frac{v}{3} \rceil$, and then add it to the priority queue. After repeating this process $k$ times, we return the answer.

The time complexity is $O(n + k \times \log n)$, and the space complexity is $O(n)$ or $O(1)$. Here, $n$ is the length of the array $nums$.

• class Solution {
public long maxKelements(int[] nums, int k) {
PriorityQueue<Integer> pq = new PriorityQueue<>((a, b) -> b - a);
for (int v : nums) {
pq.offer(v);
}
long ans = 0;
while (k-- > 0) {
int v = pq.poll();
ans += v;
pq.offer((v + 2) / 3);
}
return ans;
}
}

• class Solution {
public:
long long maxKelements(vector<int>& nums, int k) {
priority_queue<int> pq(nums.begin(), nums.end());
long long ans = 0;
while (k--) {
int v = pq.top();
pq.pop();
ans += v;
pq.push((v + 2) / 3);
}
return ans;
}
};

• class Solution:
def maxKelements(self, nums: List[int], k: int) -> int:
h = [-v for v in nums]
heapify(h)
ans = 0
for _ in range(k):
v = -heappop(h)
ans += v
heappush(h, -(ceil(v / 3)))
return ans


• func maxKelements(nums []int, k int) (ans int64) {
h := &hp{nums}
heap.Init(h)
for ; k > 0; k-- {
v := h.pop()
ans += int64(v)
h.push((v + 2) / 3)
}
return
}

type hp struct{ sort.IntSlice }

func (h hp) Less(i, j int) bool { return h.IntSlice[i] > h.IntSlice[j] }
func (h *hp) Push(v any)        { h.IntSlice = append(h.IntSlice, v.(int)) }
func (h *hp) Pop() any {
a := h.IntSlice
v := a[len(a)-1]
h.IntSlice = a[:len(a)-1]
return v
}
func (h *hp) push(v int) { heap.Push(h, v) }
func (h *hp) pop() int   { return heap.Pop(h).(int) }

• function maxKelements(nums: number[], k: number): number {
const pq = new MaxPriorityQueue();
nums.forEach(num => pq.enqueue(num));
let ans = 0;
while (k > 0) {
const v = pq.dequeue()!.element;
ans += v;
pq.enqueue(Math.floor((v + 2) / 3));
k--;
}
return ans;
}


• use std::collections::BinaryHeap;

impl Solution {
pub fn max_kelements(nums: Vec<i32>, k: i32) -> i64 {
let mut pq = BinaryHeap::from(nums);
let mut ans = 0;
let mut k = k;
while k > 0 {
if let Some(v) = pq.pop() {
ans += v as i64;
pq.push((v + 2) / 3);
k -= 1;
}
}
ans
}
}