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Minimum Number of Seconds to Make Mountain Height Zero

Difficulty: Medium

Problem Description

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, to decrease the mountain's height by x, it takes workerTimes[i] + workerTimes[i] * 2 + ... + workerTimes[i] * x seconds. Return an integer representing the minimum number of seconds required for the workers to make the height of the mountain 0.

Key Insights

  • Each worker takes increasing time to reduce the mountain height based on the current height being reduced.
  • Workers can work simultaneously, meaning the total time needed is determined by the worker that takes the longest given their assigned height reduction.
  • A binary search approach can be utilized to determine the minimum time required to reduce the mountain height to zero.

Space and Time Complexity

Time Complexity: O(mountainHeight * log(n)), where n is the number of workers. Space Complexity: O(1), as we are using a constant amount of extra space.

Solution

To solve the problem, we can employ a binary search strategy to find the minimum amount of time needed to reduce the mountain height to zero. The key steps are:

  1. Define a function that calculates the total height that can be reduced by all workers within a certain time limit.
  2. Use binary search to find the minimum time such that the total reduced height meets or exceeds the mountain height.

Code Solutions

def minTimeToReduceMountainHeight(mountainHeight, workerTimes):
    def canReduceInTime(time):
        total_reduced_height = 0
        for worker_time in workerTimes:
            # Calculate how many units of height this worker can reduce in 'time'
            x = 0
            while True:
                cost = worker_time * (x + 1) * (x + 2) // 2
                if cost > time:
                    break
                total_reduced_height += 1
                x += 1
        return total_reduced_height >= mountainHeight

    left, right = 1, sum(workerTimes) * mountainHeight  # Maximum time needed
    while left < right:
        mid = (left + right) // 2
        if canReduceInTime(mid):
            right = mid
        else:
            left = mid + 1

    return left
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