How do you solve Earliest Finish Time for Land and Water Rides I on LeetCode?

Earliest Finish Time for Land and Water Rides I (LeetCode #3633) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n + m log m + n + m) time complexity and O(n + m) space complexity. Key insight: Both rides must be experienced, so we need to consider both orders.

What is the time complexity of Earliest Finish Time for Land and Water Rides I?

The optimal solution for Earliest Finish Time for Land and Water Rides I runs in O(n log n + m log m + n + m) time and uses O(n + m) space. Sorting takes O(n log n) and O(m log m), while the two-pointer traversal takes O(n + m).

What is the brute force solution for Earliest Finish Time for Land and Water Rides I?

The brute force approach for Earliest Finish Time for Land and Water Rides I is Brute Force, with O(n²) time complexity and O(1) space complexity. We will try every combination of land and water rides to find the earliest finish time. This method checks all pairs, ensuring we consider both orders of rides. The optimal Optimal Solution approach improves this to O(n log n + m log m + n + m).

What algorithm or data structure is used to solve Earliest Finish Time for Land and Water Rides I?

Earliest Finish Time for Land and Water Rides I uses the following concepts: Array, Two Pointers, Binary Search, Greedy, Sorting. The recommended patterns to study are: Two Pointers, Sorting.

What are the interview tips for Earliest Finish Time for Land and Water Rides I?

Explain your thought process clearly. Discuss edge cases, like overlapping times. Practice similar problems to build confidence.

What are common mistakes when solving Earliest Finish Time for Land and Water Rides I?

Not considering the order of rides. Forgetting to wait for rides to open.

#3633

Earliest Finish Time for Land and Water Rides I

Easy

ArrayTwo PointersBinary SearchGreedySortingTwo PointersSorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n + m log m + n + m)
Space	O(1)	O(n + m)

💡

Intuition

Time O(n log n + m log m + n + m)Space O(n + m)

By sorting both ride categories and using a two-pointer technique, we can efficiently find the earliest finish time without checking every combination.

⚙️

Algorithm

3 steps

1Step 1: Create pairs of (start time, duration) for both ride types and sort them.
2Step 2: Use two pointers to traverse both sorted lists and calculate finish times.
3Step 3: Track the minimum finish time while ensuring rides are started at their opening times or later.

solution.py14 lines

1def earliestFinish(landStartTime, landDuration, waterStartTime, waterDuration):
2    land = sorted(zip(landStartTime, landDuration))
3    water = sorted(zip(waterStartTime, waterDuration))
4    min_finish_time = float('inf')
5    j = 0
6    for l_start, l_duration in land:
7        l_finish = l_start + l_duration
8        while j < len(water) and water[j][0] < l_finish:
9            j += 1
10        if j < len(water):
11            min_finish_time = min(min_finish_time, l_finish + water[j][1])
12        if j > 0:
13            min_finish_time = min(min_finish_time, max(water[j-1][0] + water[j-1][1], l_start) + l_duration)
14    return min_finish_time

ℹ

Complexity note: Sorting takes O(n log n) and O(m log m), while the two-pointer traversal takes O(n + m).

1Both rides must be experienced, so we need to consider both orders.
2Sorting helps to efficiently pair rides based on their start times.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.