Description
You are given two categories of theme park attractions: land rides and water rides.
- Land rides
landStartTime[i]– the earliest time theithland ride can be boarded.landDuration[i]– how long theithland ride lasts.
- Water rides
waterStartTime[j]– the earliest time thejthwater ride can be boarded.waterDuration[j]– how long thejthwater 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 timet + 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 at2 + landDuration[0] = 6. - Water ride 0 opens at time
waterStartTime[0] = 6. Start immediately at6, finish at6 + waterDuration[0] = 9.
- Start land ride 0 at time
- Plan B (water ride 0 → land ride 1):
- Start water ride 0 at time
waterStartTime[0] = 6. Finish at6 + waterDuration[0] = 9. - Land ride 1 opens at
landStartTime[1] = 8. Start at time9, finish at9 + landDuration[1] = 10.
- Start water ride 0 at time
- Plan C (land ride 1 → water ride 0):
- Start land ride 1 at time
landStartTime[1] = 8. Finish at8 + landDuration[1] = 9. - Water ride 0 opened at
waterStartTime[0] = 6. Start at time9, finish at9 + waterDuration[0] = 12.
- Start land ride 1 at time
- Plan D (water ride 0 → land ride 0):
- Start water ride 0 at time
waterStartTime[0] = 6. Finish at6 + waterDuration[0] = 9. - Land ride 0 opened at
landStartTime[0] = 2. Start at time9, finish at9 + landDuration[0] = 13.
- Start water ride 0 at time
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 at1 + waterDuration[0] = 11. - Land ride 0 opened at
landStartTime[0] = 5. Start immediately at11and finish at11 + landDuration[0] = 14.
- Start water ride 0 at time
- Plan B (land ride 0 → water ride 0):
- Start land ride 0 at time
landStartTime[0] = 5. Finish at5 + landDuration[0] = 8. - Water ride 0 opened at
waterStartTime[0] = 1. Start immediately at8and finish at8 + waterDuration[0] = 18.
- Start land ride 0 at time
Plan A provides the earliest finish time of 14.
Constraints:
1 <= n, m <= 5 * 104landStartTime.length == landDuration.length == nwaterStartTime.length == waterDuration.length == m1 <= landStartTime[i], landDuration[i], waterStartTime[j], waterDuration[j] <= 105
Solutions
This code calculates the earliest possible time to complete two sequential tasks—one on land and one on water—where one task must finish before the other can begin, and each task has a specific scheduled start time and duration. To find the fastest overall time, the code compares two strategies: completing a land task first, or completing a water task first. For the "land-first" strategy, it finds the absolute earliest time any land task can finish, and then determines which water task can start and finish the quickest after that moment. It then performs the reverse calculation for the "water-first" strategy, finding the earliest time any water task can finish and pairing it with the quickest subsequent land task. Finally, it returns the minimum finish time between these two strategies.
/**
* @param {number[]} landStartTime
* @param {number[]} landDuration
* @param {number[]} waterStartTime
* @param {number[]} waterDuration
* @return {number}
*/
var earliestFinishTime = function(landStartTime, landDuration, waterStartTime, waterDuration) {
const m = landStartTime.length;
const n = waterStartTime.length;
let minl = Infinity;
let minw = Infinity;
let res = Infinity;
for (let i = 0; i < m; i++) {
minl = Math.min(minl, landStartTime[i] + landDuration[i]);
}
for (let j = 0; j < n; j++) {
minw = Math.min(minw, waterStartTime[j] + waterDuration[j]);
res = Math.min(res, Math.max(minl, waterStartTime[j]) + waterDuration[j]);
}
for (let i = 0; i < m; i++) {
res = Math.min(res, Math.max(minw, landStartTime[i]) + landDuration[i]);
}
return res;
};