min_num_sec_mountain_height.py (1144B)
| # 3296. Minimum Number of Seconds to Make Mountain Height Zero
import heapq
from typing import List
"""
you are given an integer 'mountainHeight' denoting the height of a mountain.
you are also given an integer array 'workerTimes' representing the work time
of workers in seconds. the workers simultaneously work to reduce the height
of a mountain. return an integer representing the minimum number of seconds
required foor the workers to make the height of the mountain 0.
"""
class Solution:
def minNumberOfSeconds(self, mountainHeight: int, workerTimes: List[int]) -> int:
h = [(j, j, 1) for i, j in enumerate(workerTimes)]
heapq.heapify(h)
while mountainHeight > 1:
ps, wt, x = heapq.heappop(h)
heapq.heappush(h, (ps + wt * (x + 1), wt, x + 1))
mountainHeight -= 1
return heapq.heappop(h)[0]
if __name__ == "__main__":
obj = Solution()
print(obj.minNumberOfSeconds(mountainHeight=4, workerTimes=[2, 1, 1]))
print(obj.minNumberOfSeconds(mountainHeight=10, workerTimes=[3, 2, 2, 4]))
print(obj.minNumberOfSeconds(mountainHeight=5, workerTimes=[1]))
|
