Introduction

In this blog post, we will explore how to solve the LeetCode problem "3296" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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. For example: To reduce the height of the mountain by 1, it takes workerTimes[i] seconds. To reduce the height of the mountain by 2, it takes 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: Worker 0 reduces the height by 1, taking workerTimes[0] = 2 seconds. Worker 1 reduces the height by 2, taking workerTimes[1] + workerTimes[1] 2 = 3 seconds. Worker 2 reduces the height by 1, taking 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: Worker 0 reduces the height by 2, taking workerTimes[0] + workerTimes[0] 2 = 9 seconds. Worker 1 reduces the height by 3, taking workerTimes[1] + workerTimes[1] 2 + workerTimes[1] 3 = 12 seconds. Worker 2 reduces the height by 3, taking workerTimes[2] + workerTimes[2] 2 + workerTimes[2] 3 = 12 seconds. Worker 3 reduces the height by 2, taking 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 <= 105 1 <= workerTimes.length <= 104 1 <= workerTimes[i] <= 106

Explanation

Here's the solution to the problem:

• Binary Search for Minimum Time: The core idea is to use binary search to find the minimum time required. The search space is between 0 and the maximum possible time (mountainHeight * max(workerTimes)).

• Check Feasibility: For a given time t, we check if it's feasible to reduce the mountain to height 0 within that time. This involves determining how much each worker can reduce the mountain's height within the given time t and summing up the total reduction.

• Optimization of Height Reduction Calculation: We use the formula for the sum of an arithmetic series to efficiently calculate the height reduction for each worker within the given time t.

• Runtime & Storage Complexity:

  • Runtime Complexity: O(n log (m * k)), where n is the number of workers, m is the mountain height, and k is the maximum work time.
  • Storage Complexity: O(1)

Code

    def minTimeToZero(mountainHeight: int, workerTimes: list[int]) -> int:
    """
    Calculates the minimum time required for workers to reduce the mountain height to 0.

    Args:
        mountainHeight: The initial height of the mountain.
        workerTimes: A list of worker times.

    Returns:
        The minimum time required to reduce the mountain height to 0.
    """

    def check(time: int) -> bool:
        """
        Checks if it's possible to reduce the mountain height to 0 within the given time.

        Args:
            time: The time to check.

        Returns:
            True if possible, False otherwise.
        """
        total_reduction = 0
        for worker_time in workerTimes:
            # Calculate the maximum x such that worker_time * (x * (x + 1)) / 2 <= time
            # x * (x + 1) <= 2 * time / worker_time
            # x^2 + x - 2 * time / worker_time <= 0
            # x = (-1 + sqrt(1 + 8 * time / worker_time)) / 2
            max_reduction = int(((-1 + (1 + 8 * time / worker_time)**0.5) / 2))
            total_reduction += max_reduction

        return total_reduction >= mountainHeight

    low = 0
    high = mountainHeight * max(workerTimes)  # Maximum possible time
    ans = high

    while low <= high:
        mid = (low + high) // 2
        if check(mid):
            ans = mid
            high = mid - 1
        else:
            low = mid + 1

    return ans