LeetCode 3296 - Minimum Number of Seconds to Make Mountain Height Zero

The problem asks us to calculate the minimum amount of time required for a team of workers to reduce the height of a mountain to zero, given that each worker has a specific time cost for performing work and the total work time grows linearly for consecutive units of height.

leetcodemediumarraymathbinary-searchgreedyheap-(priority-queue)

LeetCode Problem 3296

Difficulty: 🟡 Medium
Topics: Array, Math, Binary Search, Greedy, Heap (Priority Queue)

Solution

Problem Understanding

The problem asks us to calculate the minimum amount of time required for a team of workers to reduce the height of a mountain to zero, given that each worker has a specific time cost for performing work and the total work time grows linearly for consecutive units of height. Specifically, if a worker i with workerTimes[i] works for x units, the total time they spend is the sum of the arithmetic series workerTimes[i] + 2*workerTimes[i] + ... + x*workerTimes[i]. All workers work simultaneously, so the total time to complete the work is the maximum time taken by any worker.

Inputs are:

  • mountainHeight (1 to 105), the total height to reduce.
  • workerTimes (length 1 to 104), an array of integers (1 to 106) representing the seconds a worker needs per unit increment of work.

The output is a single integer representing the minimum number of seconds to reduce the mountain height to zero.

Constraints indicate that a brute-force simulation of every possible assignment of mountain units to workers is not feasible due to high potential input sizes. Key edge cases include having only one worker, where the worker must perform all work, or multiple workers with varying speeds, where distributing work optimally is crucial.

Approaches

Brute Force

A brute-force approach would try all distributions of mountain height among the workers and compute the maximum time for each configuration. While it guarantees correctness, this is impractical because the number of distributions grows combinatorially with mountainHeight and the number of workers. For example, mountainHeight = 10^5 and len(workerTimes) = 10^4 makes brute-force impossible.

Optimal Approach

The key insight is that the time taken by a worker for x units of work is the arithmetic sum workerTimes[i] * (1 + 2 + ... + x) = workerTimes[i] * x * (x + 1) / 2. This is a monotonic function of x. Since we want to minimize the maximum time across workers, we can use binary search on the total time T. For a candidate T, we calculate the maximum height each worker can reduce within T seconds using the formula x = floor((-1 + sqrt(1 + 8*T/workerTimes[i])) / 2). Summing these heights across all workers, if it reaches mountainHeight, the time T is feasible. Binary search efficiently finds the minimum feasible T.

Approach Time Complexity Space Complexity Notes
Brute Force O((mountainHeight)^(len(workerTimes))) O(1) Tries all distributions of mountain units among workers
Optimal (Binary Search + Math) O(n * log(maxTime)) O(1) Binary search on time with arithmetic series calculation

Algorithm Walkthrough

  1. Initialize binary search bounds. The lower bound is 1, and the upper bound is mountainHeight * max(workerTimes) * (mountainHeight + 1) / 2, which is the maximum time a single worker could take if they did all the work.
  2. While the search interval is valid, pick mid = (low + high) // 2 as the candidate total time.
  3. For each worker, compute the maximum height they can reduce in mid seconds using x = floor((-1 + sqrt(1 + 8*mid/workerTimes[i])) / 2).
  4. Sum the heights across all workers.
  5. If the sum is greater than or equal to mountainHeight, mid is feasible. Set high = mid - 1.
  6. Otherwise, mid is too small. Set low = mid + 1.
  7. Continue until low > high. The minimum feasible time is low.

Why it works: The maximum units a worker can reduce in a given time is strictly increasing with time. Binary search exploits this monotonicity, and the sum across workers guarantees feasibility. This ensures the algorithm finds the minimum total time.

Python Solution

from typing import List
import math

class Solution:
    def minNumberOfSeconds(self, mountainHeight: int, workerTimes: List[int]) -> int:
        def max_height(time: int) -> int:
            total = 0
            for wt in workerTimes:
                # Solve k*(k+1)/2 * wt <= time
                k = int((-1 + math.isqrt(1 + 8 * time // wt)) // 1)
                total += k
            return total
        
        left, right = 1, mountainHeight * max(workerTimes) * (mountainHeight + 1) // 2
        answer = right
        while left <= right:
            mid = (left + right) // 2
            if max_height(mid) >= mountainHeight:
                answer = mid
                right = mid - 1
            else:
                left = mid + 1
        return answer

This implementation defines max_height(time) to calculate how many mountain units can be reduced in time. Binary search then narrows down the minimum feasible total time.

Go Solution

import (
    "math"
)

func minNumberOfSeconds(mountainHeight int, workerTimes []int) int64 {
    maxHeight := func(time int64) int64 {
        var total int64
        for _, wt := range workerTimes {
            t := int64(wt)
            k := int64(math.Floor((-1 + math.Sqrt(1 + 8*float64(time)/float64(t))) / 2))
            total += k
        }
        return total
    }

    left := int64(1)
    right := int64(mountainHeight) * int64(maxInt(workerTimes)) * int64(mountainHeight+1) / 2
    var answer int64 = right
    for left <= right {
        mid := (left + right) / 2
        if maxHeight(mid) >= int64(mountainHeight) {
            answer = mid
            right = mid - 1
        } else {
            left = mid + 1
        }
    }
    return answer
}

func maxInt(arr []int) int {
    m := arr[0]
    for _, v := range arr {
        if v > m {
            m = v
        }
    }
    return m
}

Go implementation mirrors Python logic, with careful casting to int64 to avoid integer overflow, and uses math.Sqrt for square root calculations.

Worked Examples

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

Worker Max Height in 3 sec
2 1
1 2
1 1

Sum = 4 ≥ 4, feasible. Minimum time = 3.

Example 2: mountainHeight = 10, workerTimes = [3,2,2,4]

Binary search finds time = 12. Heights assigned to workers: 2,3,3,2. Maximum time = 12, sum ≥ 10.

Example 3: mountainHeight = 5, workerTimes = [1]

Only one worker: time = 1+2+3+4+5 = 15. Minimum time = 15.

Complexity Analysis

Measure Complexity Explanation
Time O(n * log(m * H^2)) Binary search over total time, O(log(max possible time)) iterations, n workers per check
Space O(1) Only counters and bounds are stored, no extra structures

The time bound arises because maximum time could be mountainHeight * max(workerTimes) * (mountainHeight + 1)/2, and each iteration sums over all workers.

Test Cases

# Provided examples
assert Solution().minNumberOfSeconds(4, [2,1,1]) == 3  # multiple workers, small height
assert Solution().minNumberOfSeconds(10, [3,2,2,4]) == 12  # multiple workers, larger height
assert Solution().minNumberOfSeconds(5, [1]) == 15  # single worker

# Edge cases
assert Solution().minNumberOfSeconds(1, [1]) == 1  # smallest input
assert Solution().minNumberOfSeconds(100000, [1]) == 5000050000  # single worker, large height
assert Solution().minNumberOfSeconds(100000, [1000000]) == 5000050000  # very fast worker, large height
assert Solution().minNumberOfSeconds(5, [1,1,1,1,1]) == 2  # multiple slow workers, small height
Test Why
(4, [2,1,1]) Validates multiple workers with varying speeds
(10, [3,2,2,4]) Validates distribution across multiple workers
(5, [1]) Single worker case
(1, [1]) Minimum possible input
(100000, [1]) Stress test with single worker and large height
(5, [1,1,1,1,1]) Tests parallelism with multiple workers

Edge Cases

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