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Tested Python solution for LeetCode 2530 with 17 pytest cases. Generate a practice environment with lcpy.
lcpy gen -n 2530 # by problem number
lcpy gen -s maximal_score_after_applying_k_operations # by problem name
nums and an integer k. You have a starting score of 0.
In one operation:
i such that 0 <= i < nums.length,nums[i], andnums[i] with ceil(nums[i] / 3).k operations.
The ceiling function ceil(val) is the least integer greater than or equal to val.
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.
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 [1,2,1,3,3]. Your score increases by 3.
The final score is 10 + 4 + 3 = 17.
import heapq
class Solution:
# Time: O(n + k * log n)
# Space: O(n)
def max_kelements(self, nums: list[int], k: int) -> int:
heap = [-num for num in nums]
heapq.heapify(heap)
score = 0
for _ in range(k):
num = -heapq.heappop(heap)
score += num
heapq.heappush(heap, -((num + 2) // 3))
return score
| Time | Space |
|---|---|
| O(n + k * log n) | O(n) |