#1883

Hard

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Minimum Skips to Arrive at Meeting On Time

Solve the problem of minimizing skips while traveling to arrive on time, using dynamic programming and state transitions.

array dynamic programming

Problem Statement

You have a meeting to attend and need to travel through multiple roads to reach it. The roads are described by an array dist where each element represents the length of a road in kilometers. You have hoursBefore hours to reach your meeting, and you travel at a constant speed. After traveling each road, you are required to rest for one hour before continuing, except for the last road. However, you are allowed to skip some of the rest periods in order to reach your meeting on time. Your task is to determine the minimum number of skips needed, or return -1 if it’s impossible to arrive on time.

The travel time for each road is determined by dividing the road length by your speed. Skipping a rest shortens the total time, allowing you to reach the destination earlier. The challenge is to calculate how many rests you can skip while ensuring you stay within the allowed travel time, taking into account the impact of skipping each rest on your total time.

Examples

Example 1

Input: dist = [1,3,2], speed = 4, hoursBefore = 2

Output: 1

Without skipping any rests, you will arrive in (1/4 + 3/4) + (3/4 + 1/4) + (2/4) = 2.5 hours. You can skip the first rest to arrive in ((1/4 + 0) + (3/4 + 0)) + (2/4) = 1.5 hours. Note that the second rest is shortened because you finish traveling the second road at an integer hour due to skipping the first rest.

Example 2

Input: dist = [7,3,5,5], speed = 2, hoursBefore = 10

Output: 2

Without skipping any rests, you will arrive in (7/2 + 1/2) + (3/2 + 1/2) + (5/2 + 1/2) + (5/2) = 11.5 hours. You can skip the first and third rest to arrive in ((7/2 + 0) + (3/2 + 0)) + ((5/2 + 0) + (5/2)) = 10 hours.

Example 3

Input: dist = [7,3,5,5], speed = 1, hoursBefore = 10

Output: -1

It is impossible to arrive at the meeting on time even if you skip all the rests.

Constraints

n == dist.length
1 <= n <= 1000
1 <= dist[i] <= 105
1 <= speed <= 106
1 <= hoursBefore <= 107

Solution Approach

Dynamic Programming with State Transitions

Use dynamic programming (DP) to maintain a state that tracks the time needed to reach the end of each road while considering the number of skips. At each step, calculate the time for each road, considering whether or not to skip the rest at that point. Update the DP table with the minimum number of skips required for each possible state.

Minimizing Skips by Checking Feasibility

At each road, consider whether skipping the rest is feasible by comparing the remaining time to the required time to complete the journey. If skipping is not possible, proceed with the standard rest. This allows for a dynamic calculation of the skips needed to stay on track.

Edge Case Handling

Ensure that edge cases like being unable to arrive on time even with no skips, or having only one road, are handled. If the total time exceeds the available hours before the meeting, return -1. Otherwise, track the number of skips and return the minimum number.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time and space complexity depend on the final approach, but with dynamic programming, the time complexity is typically O(n * k), where n is the number of roads and k is the number of skips. The space complexity is O(n * k) due to the DP table that stores the state for each road and possible skip count.

What Interviewers Usually Probe

Is the candidate familiar with dynamic programming and state transitions?
Can the candidate effectively handle edge cases and constraints, such as when it's impossible to arrive on time?
Does the candidate consider both the travel time and the skips when formulating the solution?

Common Pitfalls or Variants

Common pitfalls

Forgetting to account for the last road not requiring a rest.
Misunderstanding the calculation of travel time for each road and the impact of skips.
Overcomplicating the DP solution or failing to optimize the state transitions.

Follow-up variants

Modify the problem by introducing variable speeds for different roads.
Change the constraint to allow skipping rests at specific times only.
Add extra conditions such as required stops for rest at specific intervals.

FAQ

How do I minimize skips in the Minimum Skips to Arrive at Meeting On Time problem?

To minimize skips, use dynamic programming to track the minimum number of skips required to arrive on time, considering the travel time and possible skips for each road.

What dynamic programming approach is needed for this problem?

The approach involves maintaining a DP table that tracks the current state at each road, considering both the time and the number of skips needed.

What is the time complexity of the Minimum Skips to Arrive at Meeting On Time problem?

The time complexity depends on the final approach, typically O(n * k), where n is the number of roads and k is the maximum number of skips.

Can the solution be optimized for larger inputs?

Yes, the solution can be optimized by refining the DP approach and reducing unnecessary calculations during state transitions.

What if I cannot arrive on time even with all skips?

If it's impossible to arrive on time, the solution should return -1 to indicate the infeasibility of the journey.

terminal

Solution

Solution 1: Dynamic Programming

We define $f[i][j]$ as the shortest time considering the first $i$ roads and exactly skipping $j$ rest times. Initially, $f[0][0]=0$, and the rest $f[i][j]=\infty$.

class Solution:
    def minSkips(self, dist: List[int], speed: int, hoursBefore: int) -> int:
        n = len(dist)
        f = [[inf] * (n + 1) for _ in range(n + 1)]
        f[0][0] = 0
        eps = 1e-8
        for i, x in enumerate(dist, 1):
            for j in range(i + 1):
                if j < i:
                    f[i][j] = min(f[i][j], ceil(f[i - 1][j] + x / speed - eps))
                if j:
                    f[i][j] = min(f[i][j], f[i - 1][j - 1] + x / speed)
        for j in range(n + 1):
            if f[n][j] <= hoursBefore + eps:
                return j
        return -1

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