Solution Approach

The implementation uses binary search to find the minimum speed required. Let's walk through the key components:

Initial Feasibility Check: First, we check if it's possible to reach on time. If the number of trains len(dist) is greater than ceil(hour), we immediately return -1. This is because even at infinite speed, we'd need at least n-1 full hours for the first n-1 trains due to the waiting constraint.

Speed Validation Function: The can_reach_in_time(speed) function determines if a given speed allows us to arrive within the time limit:

def can_reach_in_time(speed: int) -> bool:
    total_time = 0
    for i, d in enumerate(dist):
        travel_time = d / speed
        if i == len(dist) - 1:
            total_time += travel_time
        else:
            total_time += ceil(travel_time)
    return total_time <= hour

For each train ride:

• Calculate travel time: travel_time = d / speed
• If it's the last train, add the exact time
• Otherwise, add ceil(travel_time) to account for waiting until the next integer hour
• Return True if total time is within the limit

Binary Search Using the Standard Template: The feasibility condition creates a pattern as speed increases:

false, false, false, true, true, true, ...

We use the template to find the first true index (minimum speed that works):

left, right = 1, 10**7
first_true_index = -1

while left <= right:
    mid = (left + right) // 2
    if can_reach_in_time(mid):  # feasible condition
        first_true_index = mid
        right = mid - 1
    else:
        left = mid + 1

Result Handling:

• If first_true_index == -1, no valid speed exists within our constraints, return -1
• Otherwise, return first_true_index as the minimum required speed

The binary search efficiently narrows down from a potential range of 10^7 values to find the exact minimum speed in O(log(10^7)) iterations, with each iteration taking O(n) time to validate the speed.

Example Walkthrough

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

Example: dist = [1, 3, 2], hour = 2.5

Step 1: Initial Feasibility Check

• Number of trains: n = 3
• ceil(2.5) = 3
• Since 3 ≤ 3, it's potentially possible to complete the journey

Step 2: Binary Search Setup

• Left boundary: l = 1 km/h
• Right boundary: r = 10^7 + 1 km/h
• We'll search for the minimum speed where we can complete the journey in ≤ 2.5 hours

Step 3: Binary Search Process

Let's trace through a few iterations:

First iteration: mid = (1 + 10^7) / 2 ≈ 5,000,000 km/h

• Train 1: 1/5,000,000 hours ≈ 0, wait until hour 1
• Train 2: 3/5,000,000 hours ≈ 0, wait until hour 2
• Train 3: 2/5,000,000 hours ≈ 0
• Total time ≈ 2 hours < 2.5 ✓
• This speed works, search lower half

Several iterations later: mid = 3 km/h

• Train 1: 1/3 = 0.33 hours → ceil(0.33) = 1 hour (wait until hour 1)
• Train 2: 3/3 = 1 hour → ceil(1) = 1 hour (depart at hour 2)
• Train 3: 2/3 = 0.67 hours (no waiting for last train)
• Total time = 1 + 1 + 0.67 = 2.67 hours > 2.5 ✗
• This speed doesn't work, search higher speeds

Next iteration: mid = 4 km/h

• Train 1: 1/4 = 0.25 hours → ceil(0.25) = 1 hour
• Train 2: 3/4 = 0.75 hours → ceil(0.75) = 1 hour
• Train 3: 2/4 = 0.5 hours
• Total time = 1 + 1 + 0.5 = 2.5 hours = 2.5 ✓
• This speed works exactly!

Continue searching: Check if speed 3 works (it doesn't, as shown above)

Step 4: Result The binary search converges to find that the minimum speed is 4 km/h.

At this speed:

• We board train 1 at hour 0, arrive at 0.25, wait until hour 1
• We board train 2 at hour 1, arrive at 1.75, wait until hour 2
• We board train 3 at hour 2, arrive at 2.5
• Total journey time: exactly 2.5 hours

This demonstrates how the binary search efficiently finds the minimum speed by testing different values and narrowing the search range based on whether each speed allows us to complete the journey on time.

Time and Space Complexity

The time complexity is O(n × log M), where n is the length of the dist array (number of train trips) and M is the upper bound of the speed (10^7 + 1 in this case).

This complexity arises from:

• The binary search performed by bisect_left over the range [1, r) where r = 10^7 + 1, which takes O(log M) iterations
• For each iteration of the binary search, the check function is called, which iterates through all n elements in the dist array, taking O(n) time
• Therefore, the total time complexity is O(n × log M)

The space complexity is O(1) as the algorithm only uses a constant amount of extra space for variables like s, t, ans, and r, regardless of the input size. The check function and binary search operation do not require any additional space that scales with the input.

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

Common Pitfalls

1. Using the wrong binary search template variant

A common mistake is using while left < right with right = mid instead of the standard template.

Wrong approach:

while left < right:
    mid = (left + right) // 2
    if can_reach_in_time(mid):
        right = mid
    else:
        left = mid + 1
return left

Correct approach (using the standard template):

first_true_index = -1
while left <= right:
    mid = (left + right) // 2
    if can_reach_in_time(mid):
        first_true_index = mid
        right = mid - 1
    else:
        left = mid + 1
return first_true_index

2. Incorrect Handling of the Last Train's Travel Time

Pitfall: A common mistake is applying the ceiling function to ALL train travel times, including the last one.

# INCORRECT implementation
def can_reach_in_time(speed: int) -> bool:
    total = 0
    for d in dist:
        total += ceil(d / speed)  # Wrong: applies ceiling to all trains
    return total <= hour

Why it's wrong: The last train doesn't require waiting since you've reached your destination. If the last train takes 0.5 hours, the total time should include exactly 0.5 hours, not 1 hour.

Solution: Always check if you're on the last train and handle it differently:

def can_reach_in_time(speed: int) -> bool:
    total = 0
    for i, d in enumerate(dist):
        t = d / speed
        total += t if i == len(dist) - 1 else ceil(t)
    return total <= hour

3. Integer Division Instead of Float Division

Pitfall: Using integer division (//) when calculating travel time, which loses precision:

# INCORRECT - loses decimal precision
travel_time = distance // speed  # Wrong: integer division

Why it's wrong: If distance is 3 and speed is 2, integer division gives 1 instead of 1.5.

Solution: Always use float division (/) to maintain precision:

travel_time = distance / speed  # Correct: float division

4. Not Checking Early Impossibility Cases

Pitfall: Performing binary search without first checking if the problem is solvable:

# INCORRECT - missing feasibility check
def minSpeedOnTime(self, dist: List[int], hour: float) -> int:
    # Directly starts binary search without checking if solution exists
    # ...

Why it's wrong: If you have 5 trains but only 3.5 hours, it's impossible to complete the journey regardless of speed, since you need at least 4 hours (waiting for integer hours for the first 4 trains).

Solution: Check if len(dist) > ceil(hour) before starting the search:

if len(dist) > ceil(hour):
    return -1