#3296

Medium

auto_awesomeBinary search over the valid answer space

LeetCode Problem Workspace

Minimum Number of Seconds to Make Mountain Height Zero

Determine the minimum seconds required to reduce a mountain to zero height using simultaneous workers efficiently.

array math binary search greedy heap priority queue

Problem Statement

You are given an integer mountainHeight representing the height of a mountain and an array workerTimes where each element indicates the time in seconds a worker takes to reduce one unit of height. Multiple workers can work simultaneously, and each worker can work continuously until the mountain reaches zero.

Your task is to compute the minimum number of seconds required to reduce the mountain to zero height. The solution should account for all workers working together efficiently and use the optimal strategy to minimize the total time.

Examples

Example 1

Input: mountainHeight = 4, workerTimes = [2,1,1]

Output: 3

One way the height of the mountain can be reduced to 0 is: 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

The number of seconds needed is max(9, 12, 12, 12) = 12 seconds.

Example 3

Input: mountainHeight = 5, workerTimes = [1]

Output: 15

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

Solution Approach

Binary Search over Answer Space

Apply binary search on the range from 1 to a large upper bound representing maximum possible time. At each mid value, check if workers can collectively reduce the mountain to zero within that time. Narrow the range until the minimum feasible time is identified.

Check Feasibility with Greedy Distribution

For each candidate time, compute the total units each worker can remove and sum them. If the sum meets or exceeds mountainHeight, the candidate time is feasible. This greedy check ensures correctness while binary search efficiently narrows down the answer.

Optimize with Array and Math Operations

Precompute work contributions using math formulas to avoid repeated loops. Sort or prioritize workers if needed to reduce checks. Efficient array handling ensures the approach scales for large workerTimes arrays and high mountainHeight values.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity is O(n * log(maxTime)) where n is the number of workers and maxTime is the upper bound for binary search. Space complexity is O(1) extra if calculations are done in-place, otherwise O(n) for storing contributions.

What Interviewers Usually Probe

Are you thinking about iterating over all possible seconds or can we use a smarter approach?
Consider how the workers' different times affect the total reduction and feasibility check.
Can you implement a binary search over the valid answer space instead of naive simulation?

Common Pitfalls or Variants

Common pitfalls

Assuming sequential work instead of simultaneous worker contributions can overestimate time.
Neglecting to check if total units reduced in candidate time meets the mountainHeight.
Failing to set a sufficiently large upper bound for binary search may give incorrect results.

Follow-up variants

Instead of finding minimum seconds, compute the maximum mountainHeight that can be reduced in a fixed number of seconds.
Consider a version where workers have cooldown periods between units of work.
Modify the problem to include varying efficiencies depending on current mountainHeight levels.

FAQ

What is the best way to approach Minimum Number of Seconds to Make Mountain Height Zero?

Use binary search over the total time and verify feasibility with a greedy sum of worker contributions.

Can we solve this problem without binary search?

A naive simulation is possible but inefficient; binary search ensures optimal time complexity and handles large inputs.

How do workerTimes affect the solution?

Workers with smaller times can reduce more units per second, and the feasibility check must sum all contributions to meet mountainHeight.

What is a common mistake when implementing this solution?

Overlooking simultaneous work and summing contributions incorrectly can lead to wrong minimum seconds.

How does the binary search pattern apply here?

The pattern searches the smallest time where the combined work of all workers can reduce the mountainHeight to zero, efficiently narrowing the range.

terminal

Solution

Solution 1: Binary Search

We notice that if all workers can reduce the mountain height to $0$ in $t$ seconds, then for any $t' > t$, the workers can also reduce the mountain height to $0$ in $t'$ seconds. Therefore, we can use binary search to find the minimum $t$ such that the workers can reduce the mountain height to $0$ in $t$ seconds.

1

2

3

4

5

6

7

8

9

class Solution:
    def minNumberOfSeconds(self, mountainHeight: int, workerTimes: List[int]) -> int:
        def check(t: int) -> bool:
            h = 0
            for wt in workerTimes:
                h += int(sqrt(2 * t / wt + 1 / 4) - 1 / 2)
            return h >= mountainHeight

        return bisect_left(range(10**16), True, key=check)

Continue Practicing

#3049 earliest second to mark indices ii #3605 minimum stability factor of array #2967 minimum cost to make array equalindromic