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Earliest Finish Time for Land and Water Rides I
Find the earliest time a tourist can finish one land and one water ride in a theme park.
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Easy · Earliest Finish Time for Land and Water Rides I core interview pattern
Answer-first summary
Find the earliest time a tourist can finish one land and one water ride in a theme park.
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The problem involves calculating the earliest finish time for two types of theme park rides. By considering all possible pairs of land and water rides, you can determine the earliest time. A brute force approach checks each possible combination to return the optimal solution.
Problem Statement
You are given two categories of theme park attractions: land rides and water rides. A tourist must experience exactly one ride from each category, in any order. Your task is to find the earliest possible time at which the tourist can finish both rides, given the start times and durations for each ride.
To solve this, you can either start with a land ride or a water ride. The start and finish times for each ride are given, and your goal is to determine the earliest finish time by considering all possible ride combinations. The challenge is to explore all combinations efficiently.
Examples
Example 1
Input: landStartTime = [2,8], landDuration = [4,1], waterStartTime = [6], waterDuration = [3]
Output: 9
Plan A gives the earliest finish time of 9.
Example 2
Input: landStartTime = [5], landDuration = [3], waterStartTime = [1], waterDuration = [10]
Output: 14
Plan A provides the earliest finish time of 14.
Constraints
- 1 <= n, m <= 100
- landStartTime.length == landDuration.length == n
- waterStartTime.length == waterDuration.length == m
- 1 <= landStartTime[i], landDuration[i], waterStartTime[j], waterDuration[j] <= 1000
Solution Approach
Brute Force Approach
By checking every possible pairing of land and water rides, we can calculate the finish times for both combinations: starting with a land ride or starting with a water ride. This method guarantees the correct result but is computationally intensive.
Sorting for Optimization
Sort the land and water rides by their start times and try to find the earliest finish time by considering each possible start combination. This reduces unnecessary calculations compared to the brute force method.
Greedy Strategy
A greedy approach can be applied by iterating over the rides, choosing the optimal next ride that minimizes the finish time based on the previous ride's completion time. This strategy improves efficiency compared to brute force.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
The brute force approach has a time complexity of O(n * m), where n is the number of land rides and m is the number of water rides. Optimizing with sorting reduces this to O(n log n + m log m), while the greedy approach operates in O(n + m) time for iterating over the rides.
What Interviewers Usually Probe
- Can the candidate identify when brute force might be too slow for larger inputs?
- Does the candidate understand the trade-offs between brute force and optimization techniques like sorting or greedy algorithms?
- Is the candidate able to balance time complexity and correctness in their approach?
Common Pitfalls or Variants
Common pitfalls
- Forgetting to consider both possible orders of land and water rides.
- Not optimizing the brute force solution by using sorting or other strategies.
- Overlooking edge cases such as having only one ride in each category.
Follow-up variants
- What if there are multiple water rides and land rides with overlapping start times?
- How would the solution change if a fixed order of rides (land first or water first) is required?
- What if you need to account for ride maintenance that causes some rides to be unavailable at certain times?
FAQ
What is the optimal solution for 'Earliest Finish Time for Land and Water Rides I'?
The optimal solution can be achieved by using sorting to reduce unnecessary brute force comparisons, improving efficiency.
How do I approach the problem if brute force is too slow?
You can optimize by sorting the rides and applying a greedy or other efficient approach to minimize the finish time.
Can I solve 'Earliest Finish Time for Land and Water Rides I' with a greedy algorithm?
Yes, a greedy approach can help by selecting the best ride at each step, minimizing the total time.
How do I deal with edge cases in 'Earliest Finish Time for Land and Water Rides I'?
Ensure that all possible ride combinations are considered, including cases where there is only one ride in each category.
What is the time complexity of the brute force approach?
The brute force approach has a time complexity of O(n * m), where n is the number of land rides and m is the number of water rides.
Solution
Solution 1: Enumeration + Greedy
We can consider two orders of rides: first land rides then water rides, or first water rides then land rides.
class Solution:
def earliestFinishTime(self, landStartTime: List[int], landDuration: List[int], waterStartTime: List[int], waterDuration: List[int]) -> int:
def calc(a1, t1, a2, t2):
min_end = min(a + t for a, t in zip(a1, t1))
return min(max(a, min_end) + t for a, t in zip(a2, t2))
x = calc(landStartTime, landDuration, waterStartTime, waterDuration)
y = calc(waterStartTime, waterDuration, landStartTime, landDuration)
return min(x, y)