3296. Minimum Number of Seconds to Make Mountain Height Zero

MediumGreedyArrayMathBinary SearchHeap (Priority Queue)

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.
- 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.

Your task is to return an integer representing the minimum number of seconds required for the workers to make the height of the mountain 0.

Intuition

To solve the problem, we need to find a way to calculate the minimum time required for workers, working simultaneously, to completely reduce the mountain height. The key observation is that if the workers can reduce the mountain height to 0 in t seconds, then they can certainly do it in more than t seconds. This creates a monotonic relationship, which allows us to use binary search over the number of seconds t to find the minimum time.

The main challenge is to determine, for a given time t, whether the workers can completely reduce the mountain. Each worker can reduce the mountain height incrementally, and the time taken follows a quadratic pattern based on their work rate, which can be expressed mathematically.

By deriving a formula, we determine h', the potential height reduction a worker can achieve in t seconds. This is expressed as:

h' <= floor(sqrt((2 * t) / workerTimes[i] + 1/4) - 1/2)

Once we have h' for each worker, we sum up these values to verify if it meets or exceeds the mountainHeight. If so, it means the chosen t seconds is sufficient; otherwise, it isn't.

We can set a search range starting from 1 to 10^16 based on the problem constraints and the possible maximum times it might take.

Thus, using binary search, we can efficiently explore this interval to pinpoint the exact minimum time needed.

Learn more about Greedy, Math, Binary Search and Heap (Priority Queue) patterns.

Solution Approach

The solution utilizes a binary search algorithm to efficiently determine the minimum number of seconds required for the workers to reduce the mountain's height to zero. Here's a step-by-step explanation of the approach:

Binary Search Setup:
- We initiate a binary search over the number of seconds t. The lower bound l is set to 1, and the upper bound r is set to a large number, 10^16. This ensures we cover the potential maximum time required.
Define the Check Function:
- The core of our approach is a function check(t), which returns True if the workers can reduce the mountain to height 0 in t seconds and False otherwise.
- For each worker, calculate the maximum possible reduction in mountain height h' that can be achieved within t seconds using:
```
h' <= floor(sqrt((2 * t) / workerTimes[i] + 1/4) - 1/2)
```
- Iterate over all workers, summing their h' values to obtain the total height reduction h.
- If the total h is at least mountainHeight, return True from check(t).
Perform Binary Search:
- Continuously divide the interval [l, r] until l equals r. During each iteration, calculate the midpoint m and use it as an argument for check(m).
- If check(m) returns True, it implies that the current t = m is sufficient, allowing us to try for a smaller time by setting r = m.
- If check(m) returns False, the current time m is not sufficient, so adjust the lower bound to search for larger times by setting l = m + 1.
Return the Result:
- Once l equals r, the minimum seconds required is found at l, providing the answer for the problem.

This approach is efficient due to binary search's O(log n) complexity combined with the logarithmic factor required to calculate h' for each worker in O(k) where k is the number of workers. The solution effectively balances between searching for the optimal time and verifying its sufficiency using mathematical reductions.

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Example Walkthrough

Let's walk through a simplified example to illustrate the solution approach.

Suppose we have a mountain with a height of mountainHeight = 6 and three workers with workerTimes = [2, 3, 4]. The goal is to determine the minimum time t required for the workers to reduce the mountain height to zero.

Binary Search Setup:
- We set initial bounds for the binary search with l = 1 and r = 10^16.
Define the Check Function:
- Create a function check(t) to determine if the workers can complete the task in t seconds.
- For each worker i, we calculate the potential height reduction h' using the formula:
```
h' <= floor(sqrt((2 * t) / workerTimes[i] + 1/4) - 1/2)
```
Perform Check at an Initial Middle Point:
- Consider an initial midpoint, for instance, m = 15.
- Calculate h' for each worker:
  - Worker 1: workerTimes[0] = 2
    - h' = floor(sqrt((2 * 15) / 2 + 0.25) - 0.5) = 3
  - Worker 2: workerTimes[1] = 3
    - h' = floor(sqrt((2 * 15) / 3 + 0.25) - 0.5) = 2
  - Worker 3: workerTimes[2] = 4
    - h' = floor(sqrt((2 * 15) / 4 + 0.25) - 0.5) = 2
- Summing their contributions: total_h = 3 + 2 + 2 = 7, which is greater than mountainHeight = 6.
- Since check(15) returns True, decrease r to m = 15.
Narrow Down the Interval with Further Binary Search:
- Repeat the process with new midpoints, refining l and r according to whether check(m) is True or False.
- For instance, try m = 7:
  - Worker 1: h' = floor(sqrt((2 * 7) / 2 + 0.25) - 0.5) = 2
  - Worker 2: h' = floor(sqrt((2 * 7) / 3 + 0.25) - 0.5) = 1
  - Worker 3: h' = floor(sqrt((2 * 7) / 4 + 0.25) - 0.5) = 1
  - total_h = 2 + 1 + 1 = 4, less than mountainHeight, so check(7) returns False.
- As check(7) is False, adjust the lower bound: l = 8.
Continue Until Convergence:
- This process continues until l equals r. Assume after several iterations, it converges to l = 10.
Conclusion:
- The algorithm determines that the minimum time required for the workers to completely reduce the mountain is t = 10 seconds.

This method efficiently pinpoints the minimum time by leveraging binary search, ensuring a rapid solution even for large problems.

Solution Implementation

1from typing import List
2from bisect import bisect_left
3from math import sqrt
4
5class Solution:
6    def minNumberOfSeconds(self, mountainHeight: int, workerTimes: List[int]) -> int:
7        # Helper function to determine if a given amount of time 't' is sufficient
8        def check(t: int) -> bool:
9            h = 0  # This will keep track of the total height climbed
10            for wt in workerTimes:
11                # Calculate maximum height one worker can climb in time 't'
12                h += int(sqrt(2 * t / wt + 1 / 4) - 1 / 2)
13            return h >= mountainHeight  # Check if the total climbed height is sufficient
14      
15        # Perform binary search across an extensive range to minimize time
16        return bisect_left(range(10**16), True, key=check)
17

1class Solution {
2    private int mountainHeight;  // Height of the mountain to reach
3    private int[] workerTimes;   // Array storing the times each worker takes
4
5    // Method to find the minimum number of seconds required
6    public long minNumberOfSeconds(int mountainHeight, int[] workerTimes) {
7        this.mountainHeight = mountainHeight;  // Initialize the mountain height
8        this.workerTimes = workerTimes;        // Initialize the worker times array
9
10        long left = 1, right = (long) 1e16;  // Set an initial range for binary search
11
12        // Binary search to find the minimum time
13        while (left < right) {
14            long mid = (left + right) / 2;  // Calculate the midpoint of the current range
15            if (check(mid)) {
16                right = mid;  // If the current mid can satisfy the height, search for a smaller time
17            } else {
18                left = mid + 1;  // Otherwise, increase the lower bound
19            }
20        }
21      
22        return left;  // The minimum time found
23    }
24
25    // Helper method to check if a given time is sufficient to reach or exceed the mountain height
26    private boolean check(long t) {
27        long totalHeight = 0;  // Initialize the total height that can be climbed
28
29        // Calculate the total height all workers can climb in time t
30        for (int time : workerTimes) {
31            // Calculate the amount of height each worker can climb using physics formula rearrangement
32            totalHeight += (long) (Math.sqrt(t * 2.0 / time + 0.25) - 0.5);
33        }
34
35        // Return true if the total height is at least the mountain height
36        return totalHeight >= mountainHeight;
37    }
38}
39

1#include <vector>
2#include <cmath>
3using namespace std;
4
5class Solution {
6public:
7    long long minNumberOfSeconds(int mountainHeight, vector<int>& workerTimes) {
8        // Define a type alias for long long
9        using ll = long long;
10
11        // Set initial binary search bounds: l is low bound, r is high bound
12        ll l = 1, r = 1e16; 
13
14        // Lambda function to check if a given time t is sufficient
15        auto check = [&](ll t) -> bool {
16            ll accumulatedHeight = 0; // Accumulator for total height ascended by workers
17
18            // Iterate over each worker's time
19            for (int& wt : workerTimes) {
20                // Calculate the maximum height this worker can climb within time t
21                accumulatedHeight += (ll)(sqrt(t * 2.0 / wt + 0.25) - 0.5);
22            }
23
24            // Check if accumulated height is greater than or equal to mountainHeight
25            return accumulatedHeight >= mountainHeight;
26        };
27
28        // Perform binary search to find the minimum time required
29        while (l < r) {
30            ll mid = (l + r) >> 1; // Calculate the midpoint
31
32            // Use the check function to see if mid time is sufficient
33            if (check(mid)) {
34                r = mid; // If sufficient, update the upper bound
35            } else {
36                l = mid + 1; // Otherwise, update the lower bound
37            }
38        }
39
40        return l; // Return the minimum time found
41    }
42};
43

1// Function to calculate the minimum number of seconds needed to climb a given mountainHeight using workers with specified times.
2function minNumberOfSeconds(mountainHeight: number, workerTimes: number[]): number {
3
4    // Helper function to check if it's possible to accumulate enough height in given time t.
5    const check = (t: bigint): boolean => {
6        let accumulatedHeight = BigInt(0); // Initialize the total height collected
7
8        // Iterate through each worker's time and calculate the height they can contribute
9        for (const workerTime of workerTimes) {
10            const maxHeight = Math.floor(Math.sqrt((Number(t) * 2.0) / workerTime + 0.25) - 0.5);
11            accumulatedHeight += BigInt(maxHeight);
12        }
13
14        // Check if accumulated height is equal to or exceeds the required mountain height
15        return accumulatedHeight >= BigInt(mountainHeight);
16    };
17
18    let left = BigInt(1); // Lower bound of the binary search space
19    let right = BigInt(1e16); // Upper bound of the binary search space
20
21    // Binary search to find the minimum amount of time needed
22    while (left < right) {
23        const mid = (left + right) >> 1n; // Calculate the midpoint
24        if (check(mid)) {
25            right = mid; // If it's possible, narrow the search to the left half
26        } else {
27            left = mid + 1n; // If not possible, narrow the search to the right half
28        }
29    }
30
31    // Return the minimum time found, converted back to number from big integer
32    return Number(left);
33}
34

Time and Space Complexity

The time complexity of the provided code is O(n \times \log M), where n is the number of workers and M is the right boundary of the binary search, which is 10^{16} in this problem. This is because for each step of the binary search, the check function iterates over all workers to compute the total height h the workers can achieve.

The space complexity is O(1), as the algorithm uses a fixed amount of extra space regardless of the input sizes. The primary operations are calculations involving integers and a few temporary variables.

Learn more about how to find time and space complexity quickly.

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