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Minimum Number of Seconds to Make Mountain Height Zero

Number: 3496

Difficulty: Medium

Paid? No

Companies: N/A

Problem Description

Given a mountain of certain height and an array of worker times, each worker can reduce the mountain’s height sequentially. To reduce the mountain's height by x units, worker i takes workerTimes[i] + workerTimes[i] * 2 + ... + workerTimes[i] * x seconds, and all workers work simultaneously. Find the minimum number of seconds required so that the combined work of all workers reduces the mountain’s height to 0.

Key Insights

  • Each worker i takes workerTimes[i] * (1 + 2 + … + x) = workerTimes[i] * (x*(x+1)/2) seconds to reduce x units.
  • For a fixed time T, we can compute the maximum height each worker can reduce by solving workerTimes[i] * (x*(x+1)/2) <= T.
  • Use binary search on time T. For each candidate T, sum the maximum height each worker can reduce. If the total is at least the mountainHeight, T is sufficient.
  • Efficiently deriving each worker's contribution for a given T is key for the binary search to handle large input limits.

Space and Time Complexity

Time Complexity: O(N * log(T_max)), where N is the number of workers and T_max is the search space for time (derived from worst-case work scenario). Space Complexity: O(1) additional space (ignoring input storage).

Solution

To solve the problem, we use the following strategy:

  1. Use binary search on the time T, where T is the candidate answer.
  2. For a given T, for each worker i, compute the maximum height they can reduce:
    • This involves finding the largest integer x such that workerTimes[i] * (x*(x+1)/2) <= T.
    • We can find x by solving the quadratic inequality x^2 + x - (2*T)/workerTimes[i] <= 0.
    • The positive root of the quadratic equation is given by x = floor((sqrt(1 + 8*T/workerTimes[i]) - 1) / 2).
  3. Sum up these x values over all workers.
  4. If the total is at least mountainHeight, then T seconds are sufficient; otherwise, increase T.
  5. The binary search converges to the minimum T that satisfies the condition.

Key data structures include a simple loop over the workers for each binary search iteration. The critical algorithmic approach is binary search over time combined with solving a quadratic equation to determine each worker's contribution.

Code Solutions

Below are code solutions in multiple languages.

import math

def minimum_seconds(mountainHeight, workerTimes):
    # Helper function to check if mountain can be reduced within T seconds.
    def can_reduce(T):
        total_reduction = 0
        for time in workerTimes:
            # Solve time * (x*(x+1) / 2) <= T
            # => x^2 + x - (2*T)/time <= 0 
            # Compute discriminant
            disc = 1 + 8 * T / time
            x = int((math.sqrt(disc) - 1) // 2)
            total_reduction += x
            if total_reduction >= mountainHeight:
                return True
        return total_reduction >= mountainHeight

    # Binary search boundaries
    low, high = 0, max(workerTimes) * mountainHeight * (mountainHeight + 1) // 2
    while low < high:
        mid = (low + high) // 2
        if can_reduce(mid):
            high = mid
        else:
            low = mid + 1
    return low

# Example usage
print(minimum_seconds(4, [2,1,1]))  # Output: 3
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