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:

choose an index i such that 0 <= i < nums.length,
increase your score by nums[i], and
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 $nums$.

Python3JavaC++GoTypeScriptRust

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

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;
    }
};

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
    }
}

Solution 2

Python3C++Go

class Solution:
    def maxKelements(self, nums: List[int], k: int) -> int:
        for i, v in enumerate(nums):
            nums[i] = -v
        heapify(nums)
        ans = 0
        for _ in range(k):
            ans -= heapreplace(nums, -ceil(-nums[0] / 3))
        return ans

class Solution {
public:
    long long maxKelements(vector<int>& nums, int k) {
        make_heap(nums.begin(), nums.end());
        long long ans = 0;
        while (k--) {
            int v = nums[0];
            ans += v;
            pop_heap(nums.begin(), nums.end());
            nums.back() = (v + 2) / 3;
            push_heap(nums.begin(), nums.end());
        }
        return ans;
    }
};

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

type hp struct{ sort.IntSlice }

func (h hp) Less(i, j int) bool { return h.IntSlice[i] > h.IntSlice[j] }
func (hp) Push(any)             {}
func (hp) Pop() (_ any)         { return }

2530. Maximal Score After Applying K Operations

Description

Solutions

Solution 1: Priority Queue (Max Heap)

Solution 2

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