LeetCode #3633 — EASY

Earliest Finish Time for Land and Water Rides I

Build confidence with an intuition-first walkthrough focused on array fundamentals.

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The Problem

Problem Statement

You are given two categories of theme park attractions: land rides and water rides.

Land rides
- landStartTime[i] – the earliest time the i^th land ride can be boarded.
- landDuration[i] – how long the i^th land ride lasts.
Water rides
- waterStartTime[j] – the earliest time the j^th water ride can be boarded.
- waterDuration[j] – how long the j^th water ride lasts.

A tourist must experience exactly one ride from each category, in either order.

A ride may be started at its opening time or any later moment.
If a ride is started at time t, it finishes at time t + duration.
Immediately after finishing one ride the tourist may board the other (if it is already open) or wait until it opens.

Return the earliest possible time at which the tourist can finish both rides.

Example 1:

Input: landStartTime = [2,8], landDuration = [4,1], waterStartTime = [6], waterDuration = [3]

Output: 9

Explanation:

Plan A (land ride 0 → water ride 0):
- Start land ride 0 at time landStartTime[0] = 2. Finish at 2 + landDuration[0] = 6.
- Water ride 0 opens at time waterStartTime[0] = 6. Start immediately at 6, finish at 6 + waterDuration[0] = 9.
Plan B (water ride 0 → land ride 1):
- Start water ride 0 at time waterStartTime[0] = 6. Finish at 6 + waterDuration[0] = 9.
- Land ride 1 opens at landStartTime[1] = 8. Start at time 9, finish at 9 + landDuration[1] = 10.
Plan C (land ride 1 → water ride 0):
- Start land ride 1 at time landStartTime[1] = 8. Finish at 8 + landDuration[1] = 9.
- Water ride 0 opened at waterStartTime[0] = 6. Start at time 9, finish at 9 + waterDuration[0] = 12.
Plan D (water ride 0 → land ride 0):
- Start water ride 0 at time waterStartTime[0] = 6. Finish at 6 + waterDuration[0] = 9.
- Land ride 0 opened at landStartTime[0] = 2. Start at time 9, finish at 9 + landDuration[0] = 13.

Plan A gives the earliest finish time of 9.

Example 2:

Input: landStartTime = [5], landDuration = [3], waterStartTime = [1], waterDuration = [10]

Output: 14

Explanation:

Plan A (water ride 0 → land ride 0):
- Start water ride 0 at time waterStartTime[0] = 1. Finish at 1 + waterDuration[0] = 11.
- Land ride 0 opened at landStartTime[0] = 5. Start immediately at 11 and finish at 11 + landDuration[0] = 14.
Plan B (land ride 0 → water ride 0):
- Start land ride 0 at time landStartTime[0] = 5. Finish at 5 + landDuration[0] = 8.
- Water ride 0 opened at waterStartTime[0] = 1. Start immediately at 8 and finish at 8 + waterDuration[0] = 18.

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

Roadmap

Brute Force Baseline
Core Insight
Algorithm Walkthrough
Edge Cases
Full Annotated Code
Interactive Study Demo
Complexity Analysis

Step 01

Brute Force Baseline

Problem summary: You are given two categories of theme park attractions: land rides and water rides. Land rides landStartTime[i] – the earliest time the ith land ride can be boarded. landDuration[i] – how long the ith land ride lasts. Water rides waterStartTime[j] – the earliest time the jth water ride can be boarded. waterDuration[j] – how long the jth water ride lasts. A tourist must experience exactly one ride from each category, in either order. A ride may be started at its opening time or any later moment. If a ride is started at time t, it finishes at time t + duration. Immediately after finishing one ride the tourist may board the other (if it is already open) or wait until it opens. Return the earliest possible time at which the tourist can finish both rides.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Two Pointers · Binary Search · Greedy

Example 1

[2,8]
[4,1]
[6]
[3]

Example 2

[5]
[3]
[1]
[10]

Step 02

Core Insight

What unlocks the optimal approach

Use brute force

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3633: Earliest Finish Time for Land and Water Rides I
class Solution {
    public int earliestFinishTime(
        int[] landStartTime, int[] landDuration, int[] waterStartTime, int[] waterDuration) {
        int x = calc(landStartTime, landDuration, waterStartTime, waterDuration);
        int y = calc(waterStartTime, waterDuration, landStartTime, landDuration);
        return Math.min(x, y);
    }

    private int calc(int[] a1, int[] t1, int[] a2, int[] t2) {
        int minEnd = Integer.MAX_VALUE;
        for (int i = 0; i < a1.length; ++i) {
            minEnd = Math.min(minEnd, a1[i] + t1[i]);
        }
        int ans = Integer.MAX_VALUE;
        for (int i = 0; i < a2.length; ++i) {
            ans = Math.min(ans, Math.max(minEnd, a2[i]) + t2[i]);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3633: Earliest Finish Time for Land and Water Rides I
func earliestFinishTime(landStartTime []int, landDuration []int, waterStartTime []int, waterDuration []int) int {
	x := calc(landStartTime, landDuration, waterStartTime, waterDuration)
	y := calc(waterStartTime, waterDuration, landStartTime, landDuration)
	return min(x, y)
}

func calc(a1 []int, t1 []int, a2 []int, t2 []int) int {
	minEnd := math.MaxInt32
	for i := 0; i < len(a1); i++ {
		minEnd = min(minEnd, a1[i]+t1[i])
	}
	ans := math.MaxInt32
	for i := 0; i < len(a2); i++ {
		ans = min(ans, max(minEnd, a2[i])+t2[i])
	}
	return ans
}

# Accepted solution for LeetCode #3633: Earliest Finish Time for Land and Water Rides I
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)

// Accepted solution for LeetCode #3633: Earliest Finish Time for Land and Water Rides I
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3633: Earliest Finish Time for Land and Water Rides I
// class Solution {
//     public int earliestFinishTime(
//         int[] landStartTime, int[] landDuration, int[] waterStartTime, int[] waterDuration) {
//         int x = calc(landStartTime, landDuration, waterStartTime, waterDuration);
//         int y = calc(waterStartTime, waterDuration, landStartTime, landDuration);
//         return Math.min(x, y);
//     }
// 
//     private int calc(int[] a1, int[] t1, int[] a2, int[] t2) {
//         int minEnd = Integer.MAX_VALUE;
//         for (int i = 0; i < a1.length; ++i) {
//             minEnd = Math.min(minEnd, a1[i] + t1[i]);
//         }
//         int ans = Integer.MAX_VALUE;
//         for (int i = 0; i < a2.length; ++i) {
//             ans = Math.min(ans, Math.max(minEnd, a2[i]) + t2[i]);
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3633: Earliest Finish Time for Land and Water Rides I
function earliestFinishTime(
    landStartTime: number[],
    landDuration: number[],
    waterStartTime: number[],
    waterDuration: number[],
): number {
    const x = calc(landStartTime, landDuration, waterStartTime, waterDuration);
    const y = calc(waterStartTime, waterDuration, landStartTime, landDuration);
    return Math.min(x, y);
}

function calc(a1: number[], t1: number[], a2: number[], t2: number[]): number {
    let minEnd = Number.MAX_SAFE_INTEGER;
    for (let i = 0; i < a1.length; i++) {
        minEnd = Math.min(minEnd, a1[i] + t1[i]);
    }
    let ans = Number.MAX_SAFE_INTEGER;
    for (let i = 0; i < a2.length; i++) {
        ans = Math.min(ans, Math.max(minEnd, a2[i]) + t2[i]);
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.