
Problem Statement
In this problem, you must determine the minimum possible speed at which all trains must travel to ensure timely arrival at the office. You are provided with:
hour
- a floating-point number specifying the total time in hours available to complete the journey.dist
- an integer array where each entrydist[i]
represents the distance of the ith train segment in kilometers.
Since each train's departure is constrained to integer hours, this may necessitate waiting periods between train rides. For instance, if a train ride completes in 1.5 hours, a 0.5 hour wait is required to begin the next ride at the next integer hour mark. The challenge is to calculate the minimal integer train speed that allows you to arrive exactly on or before the given hour
, or determine if it is impossible (output = -1
).
This problem requires understanding of time constraints mixed with optimal speed calculation across potentially very lengthy train rides, each bounded by sequential and timed departure requirements.
Examples
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 <= 10^5
1 <= dist[i] <= 10^5
1 <= hour <= 10^9
- There will be at most two digits after the decimal point in
hour
.
Approach and Intuition
To approach the given problem effectively, let's break down the process into logical steps:
Understanding the Importance of Integer Hour Constraints:
- Every train ride that does not end on an integer hour forces a wait until the next full hour before the next train can depart.
Calculating Individual Train Travel Times:
- For a given speed
s
, the time to travel a distanced
isd / s
. - For all but the last train, the result is rounded up to the nearest integer hour.
- For the last train, the raw time is used without rounding.
- For a given speed
Summing Up Total Time:
- Add all rounded-up times for the first
n-1
trains. - Add the exact time for the last train.
- Add all rounded-up times for the first
Binary Search Optimization:
- Perform a binary search on possible speeds from 1 to a high upper bound (e.g.,
10^7
or more). - At each midpoint speed, compute the total travel time and compare it with
hour
. - Narrow the search range accordingly to find the smallest speed that satisfies the condition.
- Perform a binary search on possible speeds from 1 to a high upper bound (e.g.,
Edge Case:
- If even the maximum speed cannot meet the required
hour
, return-1
.
- If even the maximum speed cannot meet the required
This binary search method is necessary to handle large input sizes efficiently, reducing complexity from linear to logarithmic in the range of possible speeds.
Solutions
- C++
- Java
class Solution {
public:
double calculateTime(vector<int>& distances, int velocity) {
double totalTime = 0.0;
for (int index = 0; index < distances.size(); index++) {
double segmentTime = (double) distances[index] / (double) velocity;
totalTime += (index == distances.size() - 1 ? segmentTime : ceil(segmentTime));
}
return totalTime;
}
int minimumSpeed(vector<int>& distances, double hours) {
int low = 1;
int high = 10000000;
int optimalSpeed = -1;
while (low <= high) {
int middle = (low + high) / 2;
if (calculateTime(distances, middle) <= hours) {
optimalSpeed = middle;
high = middle - 1;
} else {
low = middle + 1;
}
}
return optimalSpeed;
}
};
The provided C++ code defines a solution for determining the minimum speed necessary to cover given distances within a specific time limit. The solution is encapsulated within a class called Solution
that includes two primary functions: calculateTime
and minimumSpeed
.
The
calculateTime
function accepts an array of distances and a velocity. It computes the total travel time needed to traverse each segment at the given velocity. For all segments except the last, the time is rounded up to the nearest whole number, accomplishing this rounding with theceil
function. The time for the last segment is added as is, without rounding.The
minimumSpeed
function uses a binary search approach to find the smallest integer speed between 1 and 10,000,000 that allows the total travel time computed bycalculateTime
to be less than or equal to the given time limit (hours
). The function initializes two pointers,low
andhigh
, to bound the possible speeds and adjusts these bounds based on the computed travel times. If a speed meets the time constraint, it is noted asoptimalSpeed
, and the search continues to see if a lower feasible speed exists by adjusting thehigh
pointer. If the computed time exceeds the allowed time, the search space is adjusted by moving thelow
pointer.
The logic in minimumSpeed
effectively narrows the search space, determining the minimum necessary speed more efficiently than a linear search, leveraging the properties of sorted data inherent in the speed values being sequentially incremental. This binary search ensures that the function will return the lowest possible speed that meets the requirement, or -1 if no valid speed can be found (although the given range should typically suffice for realistic inputs).
class Solution {
double calculateTime(int[] distances, int velocity) {
double totalTime = 0.0;
for (int idx = 0; idx < distances.length; idx++) {
double segmentTime = (double) distances[idx] / (double) velocity;
totalTime += (idx == distances.length - 1 ? segmentTime : Math.ceil(segmentTime));
}
return totalTime;
}
public int findMinimumSpeed(int[] distances, double hours) {
int low = 1;
int high = 10000000;
int optimalSpeed = -1;
while (low <= high) {
int mid = (low + high) / 2;
if (calculateTime(distances, mid) <= hours) {
optimalSpeed = mid;
high = mid - 1;
} else {
low = mid + 1;
}
}
return optimalSpeed;
}
}
The Java solution provided tackles the problem of determining the minimum speed required to travel given distances within a specified number of hours. The process leverages a binary search algorithm to efficiently find the optimal speed.
The
calculateTime
method computes the total travel time for an array of distances at a given speed. For each segment, except the last one, it rounds up the segment's travel time to the nearest whole number usingMath.ceil
. This adjustment is crucial for segments where rounding might affect the ability to meet the total hours.The
findMinimumSpeed
function initiates the search for the minimum speed necessary to meet the deadline as defined by thehours
parameter. The function defines a search range for the speed (from 1 to 10,000,000). Using a binary search approach:- The midpoint value of the current speed range (
mid
) is used as the trial speed. - The
calculateTime
method determines if the total travel time with this speed meets the deadline. - Based on this evaluation, adjust the search boundaries (
low
andhigh
) to either half the search range if the current speed is sufficient (i.e., reducehigh
), or increase it if more speed is needed (i.e., increaselow
). - This binary refinement continues until the lowest possible speed meeting the time constraint is found.
- The midpoint value of the current speed range (
The output from findMinimumSpeed
is the minimal speed required such that travel across all distances can be completed within or under the given hours, effectively balancing speed and time constraints using an optimized approach.
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