Minimum Speed to Arrive on Time

Medium

Asked by:

22 views

Topics:

ArraysBinary Search

You are given a floating-point number hour, representing the amount of time you have to reach the office. To commute to the office, you must take n trains in sequential order. You are also given an integer array dist of length n, where dist[i] describes the distance (in kilometers) of the i^th train ride.

Each train can only depart at an integer hour, so you may need to wait in between each train ride.

For example, if the 1^st train ride takes 1.5 hours, you must wait for an additional 0.5 hours before you can depart on the 2^nd train ride at the 2 hour mark.

Return the minimum positive integer speed (in kilometers per hour) that all the trains must travel at for you to reach the office on time, or -1 if it is impossible to be on time.

Tests are generated such that the answer will not exceed 10⁷ and hour will have at most two digits after the decimal point.

Example 1:

Input: dist = [1,3,2], hour = 6
Output: 1
Explanation: At speed 1:
- The first train ride takes 1/1 = 1 hour.
- Since we are already at an integer hour, we depart immediately at the 1 hour mark. The second train takes 3/1 = 3 hours.
- Since we are already at an integer hour, we depart immediately at the 4 hour mark. The third train takes 2/1 = 2 hours.
- You will arrive at exactly the 6 hour mark.

Example 2:

Input: dist = [1,3,2], hour = 2.7
Output: 3
Explanation: At speed 3:
- The first train ride takes 1/3 = 0.33333 hours.
- Since we are not at an integer hour, we wait until the 1 hour mark to depart. The second train ride takes 3/3 = 1 hour.
- Since we are already at an integer hour, we depart immediately at the 2 hour mark. The third train takes 2/3 = 0.66667 hours.
- You will arrive at the 2.66667 hour mark.

Example 3:

Input: dist = [1,3,2], hour = 1.9
Output: -1
Explanation: It is impossible because the earliest the third train can depart is at the 2 hour mark.

Constraints:

n == dist.length
1 <= n <= 10⁵
1 <= dist[i] <= 10⁵
1 <= hour <= 10⁹
There will be at most two digits after the decimal point in hour.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the values in the `dist` array? Can they be zero, negative, or floating-point numbers?
What is the maximum value of `hour`? Is it guaranteed to be non-negative?
If it's impossible to arrive on time, what value should I return? Is there a specific error code or flag?
Is it possible for `dist` to be empty or null?
Can `hour` be less than the number of elements in `dist`? If so, is it always impossible to arrive on time?

Brute Force Solution

Approach

To find the slowest speed that lets us arrive on time, we can try every possible speed. We start with a very slow speed, then slowly increase it, checking if we can arrive on time at each speed.

Here's how the algorithm would work step-by-step:

Start with a very low speed, like 1 mile per hour.
Calculate how long each segment of the journey would take at that speed.
Add up all the travel times to get the total time for the entire journey.
If the total travel time is less than or equal to the allowed time, we have a possible solution.
If not, increase the speed slightly, like by 1 mile per hour.
Repeat the process of calculating travel times and checking if the total time is within the limit, using the new speed.
Keep doing this, gradually increasing the speed, until you find the slowest speed that allows you to arrive on time.

Code Implementation

def min_speed_to_arrive_on_time_brute_force(distances, allowed_time):    minimum_speed = 1
    maximum_possible_speed = 10000000

    while minimum_speed <= maximum_possible_speed:
        is_possible = False

        for current_speed in range(minimum_speed, maximum_possible_speed + 1):
            total_time = 0
            for distance_value in distances:
                time_taken = distance_value / current_speed
                # Need to round up for all but the last segment
                if distance_value != distances[-1]:
                    time_taken = -(-time_taken // 1)

                total_time += time_taken

            # Check if a given speed allows arrival on time
            if total_time <= allowed_time:
                return current_speed
            else:
                minimum_speed += 1

        #Impossible to arrive on time
        return -1

Big(O) Analysis

Time Complexity

O(m * n) – The described solution uses an iterative approach, testing speeds incrementally until a valid speed is found. Let 'n' be the number of rails (elements in the array). For each speed, we iterate through all the rails to calculate the travel time, which takes O(n) time. Let 'm' be the number of different speed values we have to test. We continue increasing and rechecking the speeds until we find a solution, so the number of speeds we test is proportional to the magnitude of the correct speed, since we increase the speed by 1 at each step. In the worst case, we might test a large number of speeds before finding the minimum one that satisfies the condition. Therefore, the overall time complexity is O(m * n) where m represents the speed value at which the time condition is met and n represents the size of the array.

Space Complexity

O(1) – The described solution iteratively checks speeds and calculates travel time. The primary space used involves storing a few variables: the current speed being tested, the total travel time calculated for that speed, and potentially a counter for the number of train segments. These variables consume a constant amount of memory regardless of the number of train segments (N). Therefore, the auxiliary space complexity is O(1).

Optimal Solution

Approach

The key is to avoid checking every possible speed. We can use a technique to efficiently narrow down the range of possible speeds by repeatedly guessing and refining our estimate. This will quickly lead us to the slowest speed that allows us to arrive on time.

Here's how the algorithm would work step-by-step:

First, figure out the fastest and slowest possible speeds we could travel. The slowest would be almost zero and fastest would be some reasonably big number (like the longest travel distance) since traveling faster than that doesn't help us arrive earlier.
Guess a speed in the middle of this range.
Using that speed, calculate how long each leg of the journey will take. Remember to round up the time for all legs before the final one.
Add up all the times to get the total estimated travel time.
Compare the estimated travel time with the allowed travel time. If the estimated time is too long, it means our speed guess was too slow, so we need to look at faster speeds. If the estimated time is short enough, then our speed guess was perhaps too fast, and we can see if there's a slower possible acceptable speed.
Based on whether our guess was too high or too low, adjust the range of possible speeds. If our guess was too low, the new range starts from our guess and goes up to the previous highest speed. If it was too high, the new range starts from the previous lowest speed and ends at our guess.
Repeat the guessing and range adjustment until the range between the highest and lowest possible speed is very small. The final guess (or the lower value if we stopped when lower and upper are neighbors) is the slowest possible speed that lets you arrive on time.

Code Implementation

import math

def min_speed_to_arrive_on_time(distances, allowed_time):
    slowest_speed = 1
    fastest_speed = 10**7

    while slowest_speed < fastest_speed:
        current_speed = (slowest_speed + fastest_speed) // 2
        total_time = 0

        for i in range(len(distances) - 1):
            total_time += math.ceil(distances[i] / current_speed)

        total_time += distances[-1] / current_speed

        # If we can arrive on time, try a slower speed
        if total_time <= allowed_time:
            fastest_speed = current_speed
        # Otherwise, we need to go faster.
        else:
            slowest_speed = current_speed + 1

    # Check if it's impossible to arrive on time.
    if slowest_speed == 10**7 + 1:
        return -1

    return slowest_speed

Big(O) Analysis

Time Complexity

O(n log m) – The algorithm uses binary search to find the minimum speed. The search space for the speed is determined by a minimum and maximum possible speed, where 'm' represents the difference between the maximum and minimum possible speeds. Within each binary search iteration, we iterate through the input array of size 'n' to calculate the total travel time for the given speed. Therefore, each binary search step takes O(n) time. Since binary search reduces the search space by half in each step, it takes O(log m) steps. The overall time complexity is then O(n log m).

Space Complexity

O(1) – The algorithm uses a fixed number of variables to store the fastest and slowest speeds, the current guess, the estimated travel time, and the allowed travel time. No auxiliary data structures like arrays or hash maps are created that scale with the input size (N, the number of legs of the journey). Therefore, the space used remains constant regardless of the input, resulting in O(1) space complexity.

Edge Cases

Case	How to Handle
Empty `dist` array	Return -1 since it's impossible to calculate speed with no distances.
`hour` is zero or negative	Return -1 as arriving on time would require infinite speed, which is impossible.
All distances are zero	The binary search lower bound will be zero and could lead to division by zero or infinite loop; need to handle this by starting lower bound at 1.
`hour` is very small such that even traveling at maximum integer speed isn't fast enough	If the smallest possible `hour` is less than the number of distances, return -1.
`dist` contains very large distances leading to integer overflow during calculation of total time	Use long data types for time calculations to avoid overflow and potential incorrect speed assessment.
Binary search upper bound is unnecessarily high, leading to many inefficient iterations	Set the initial upper bound to a reasonable maximum speed, such as the maximum distance divided by the minimum allowed time increment (e.g., 0.001), or the maximum possible integer.
Floating-point precision issues during time calculation, leading to incorrect comparison with `hour`	Use a small tolerance (epsilon) when comparing the calculated time with `hour` to account for potential floating-point errors.
The calculated minimum speed leads to a total travel time infinitesimally smaller than `hour`, resulting in a slightly higher speed than necessary.	The binary search algorithm will eventually converge to the correct smallest speed, and no special handling is needed.

Empty `dist` array

How to Handle:

Return -1 since it's impossible to calculate speed with no distances.

`hour` is zero or negative

How to Handle:

Return -1 as arriving on time would require infinite speed, which is impossible.

All distances are zero

How to Handle:

The binary search lower bound will be zero and could lead to division by zero or infinite loop; need to handle this by starting lower bound at 1.

`hour` is very small such that even traveling at maximum integer speed isn't fast enough

How to Handle:

If the smallest possible `hour` is less than the number of distances, return -1.

`dist` contains very large distances leading to integer overflow during calculation of total time

How to Handle:

Use long data types for time calculations to avoid overflow and potential incorrect speed assessment.

Binary search upper bound is unnecessarily high, leading to many inefficient iterations

How to Handle:

Set the initial upper bound to a reasonable maximum speed, such as the maximum distance divided by the minimum allowed time increment (e.g., 0.001), or the maximum possible integer.

Floating-point precision issues during time calculation, leading to incorrect comparison with `hour`

How to Handle:

Use a small tolerance (epsilon) when comparing the calculated time with `hour` to account for potential floating-point errors.

The calculated minimum speed leads to a total travel time infinitesimally smaller than `hour`, resulting in a slightly higher speed than necessary.

How to Handle:

The binary search algorithm will eventually converge to the correct smallest speed, and no special handling is needed.