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 work simultaneously to reduce the height of the mountain. For worker i:
x, it takes workerTimes[i] + workerTimes[i] * 2 + ... + workerTimes[i] * x seconds. For example:
workerTimes[i] seconds.workerTimes[i] + workerTimes[i] * 2 seconds, and so on.Return an integer representing the minimum number of seconds required for the workers to make the height of the mountain 0.
Example 1:
Input: mountainHeight = 4, workerTimes = [2,1,1]
Output: 3
Explanation:
One way the height of the mountain can be reduced to 0 is:
workerTimes[0] = 2 seconds.workerTimes[1] + workerTimes[1] * 2 = 3 seconds.workerTimes[2] = 1 second.Since they work simultaneously, the minimum time needed is max(2, 3, 1) = 3 seconds.
Example 2:
Input: mountainHeight = 10, workerTimes = [3,2,2,4]
Output: 12
Explanation:
workerTimes[0] + workerTimes[0] * 2 = 9 seconds.workerTimes[1] + workerTimes[1] * 2 + workerTimes[1] * 3 = 12 seconds.workerTimes[2] + workerTimes[2] * 2 + workerTimes[2] * 3 = 12 seconds.workerTimes[3] + workerTimes[3] * 2 = 12 seconds.The number of seconds needed is max(9, 12, 12, 12) = 12 seconds.
Example 3:
Input: mountainHeight = 5, workerTimes = [1]
Output: 15
Explanation:
There is only one worker in this example, so the answer is workerTimes[0] + workerTimes[0] * 2 + workerTimes[0] * 3 + workerTimes[0] * 4 + workerTimes[0] * 5 = 15.
Constraints:
1 <= mountainHeight <= 1051 <= workerTimes.length <= 1041 <= workerTimes[i] <= 106This approach utilizes binary search to find the minimum possible time required by any worker to reduce the mountain's height to zero. We assume an upper bound on the time as the sum of individual times required if one worker does all the work. We binary search this time and check if it's feasible for all workers to collectively bring down the mountain within this time frame.
The solution defines a helper function, canReduce(), that checks if the mountain can be reduced within a given maximum time per worker using a greedy approach. The primary function minSecondsToReduce() performs binary search on the time to find this minimum feasible time.
JavaScript
Time Complexity: O(n * log(m * h)) where n is the number of workers, m is the maximum worker time, and h is the mountain height.
Space Complexity: O(1), as it uses a constant amount of space.
In this approach, a dynamic programming table is used to compute the minimum time required to reduce a mountain of a certain height using previous calculations. We iteratively build up the solution by considering each worker up to a given height.
This C++ implementation involves initializing a DP table of size mountainHeight + 1 with infinity, except for zero height, where the time is zero. We iterate for each worker's time, updating the table based on the possible reductions using cumulative times.
Java
Time Complexity: O(n * h^2), where n is the number of workers, and h is the mountain height. This can be high for large heights.
Space Complexity: O(h), where h is the mountain height.
| Approach | Complexity |
|---|---|
| Binary Search on Max Time | Time Complexity: O(n * log(m * h)) where n is the number of workers, m is the maximum worker time, and h is the mountain height. Space Complexity: O(1), as it uses a constant amount of space. |
| Dynamic Programming | Time Complexity: O(n * h^2), where n is the number of workers, and h is the mountain height. This can be high for large heights. Space Complexity: O(h), where h is the mountain height. |
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