A taxi driver knows the pick-up and drop-off locat...

Creato il: 8 febbraio 2025

Risposto usando O3 Mini da Chat01

Domanda

A taxi driver knows the pick-up and drop-off locations of people who are requesting taxi services.
All the locations are in km from the starting point.
The starting point is at 0 km.
For each km travelled by a passenger, the driver charges 1 unit of money per passenger. Moreover, the taxi itself is very fancy. Therefore some people are even willing to pay an extra tip if they get to travel in the taxi. At any point of time, the taxi can only accommodate one passenger. Determine the maximum amount the driver can earn.
Example
pickup = [0, 2, 9, 10, 11, 12]
drop = [5, 9, 11, 11, 14, 17)
tip = [1, 2, 3, 2, 2, 1]
The way to earn the most money is by accepting passengers at indices 1, 2 and 5.
• The amount paid by the passenger at index 1:9 - 2 +
2=9
• The amount paid by the passenger at index 2: 11 - 9

  • 3 = 5
    • The amount paid by the passenger at index 5: 17 - 12
  • 1 =6
    • The total amount paid by the passengers is 9 + 5 + 6
    = 20
    Therefore, the return value is 20.
    Function Description
    Complete the function taxiDriver in the editor below.
    The function must return an integer denoting the maximum amount that can be earned by the driver.
    taxiDriver has the following parameters):
    pickup[pickup[0]....pickup[n-1]]: an array of n integers that denote the pickup location of the potential riders
    drop[drop[0),...drop[n-1]): an array of n integers that denote the drop-off locations of the potential riders tip[tip[0],..tip[n-1]): an array of n integers that denote the tips offered by each person if they are accepted for a ride
    Constraints
    Rank
    Confident
    • 0 < | pickup|. |drop|. | tip| ≤ 104
    • 0 ≤ pickupli], dropli) ≤ 1012
    • pickup[i ] < drop(
    • 0 ≤ tip ≤ 105

in java

Pensiero

Understanding the problem

Alright, so the problem involves a taxi driver with trips where each trip has three values: pickup, drop, and tip, all of which are arrays. The driver starts at 0 km and can pick up a passenger once their pickup location is reached, then drop them off when they hit the drop location. The goal is to determine the maximum amount the driver can earn. The correct understanding seems to be that the driver earns 1 unit of money per km per passenger, plus the tips.

Breaking down the solution

Okay, I see now. The task is to maximize the driver's earnings by picking non-overlapping rides. For each ride, the earning is the distance plus the tip. The problem seems to fit into a weighted interval scheduling problem, where each ride has a start (pickup), end (drop), and weight (distance + tip). The taxi can only pick up a ride once it reaches the pickup location, and rides must be chosen such that they don’t overlap in time. Let's focus on proper sorting by drop location to handle this efficiently.

Developing the solution

The problem is essentially about maximizing earnings through weighted interval scheduling. I need to consider the recurrence: dp[i] = max(dp[i-1], dp[j] + profit[i]) where j is the ride with the rightmost drop time ≤ current ride’s pickup time.

Since pickup and drop distances can be as large as 10^12, iterating through all km isn’t feasible. Using binary search on the sorted rides array should help manage this. After sorting rides by drop time, I can apply dynamic programming to maximize the total profit.

I'll implement the solution using the dp array and binary search to find optimal rides.

Reasoned for 13 seconds

Risposta

We can solve this problem by realizing that it is a weighted interval scheduling problem. Each ride can be seen as an interval [pickupi,dropi][pickup_i, drop_i] with a profit (or weight) given by

profiti=(dropipickupi)+tipi.\text{profit}_i = (\text{drop}_i - \text{pickup}_i) + \text{tip}_i.

Because the taxi can only serve one passenger at a time (i.e. the rides cannot overlap), we want to select a set of non‐overlapping rides that maximizes the sum of profits.

A common way to solve weighted interval scheduling is as follows:

  1. Sort the rides by their drop-off location.

  2. For each ride, use binary search to find the last ride that ends at or before the current ride’s pickup. Let’s call its index jj. This tells us that if we take the current ride, then we can add its profit to the best answer obtained up to ride jj.

  3. Use dynamic programming (DP) to choose whether to take or skip a ride. If we let dp[i] be the maximum earning possible using the first ii rides (when the rides are sorted by drop), then the recurrence is

    dp[i]=max(dp[i1], dp[j+1]+profit of ride i1)dp[i] = \max(dp[i-1],\ dp[j+1] + \text{profit of ride } i-1)

    (Here we use j+1j+1 because we use a DP array of size n+1n+1 with dp[0]=0.)

Below is one acceptable Java solution:


java
import java.util.Arrays; public class TaxiDriver { // Helper class to hold ride information. static class Ride { long pickup; long drop; long profit; // profit = (drop - pickup) + tip public Ride(long pickup, long drop, int tip) { this.pickup = pickup; this.drop = drop; this.profit = (drop - pickup) + tip; } } /** * Given arrays of pickup locations, drop-off locations, and tips, * returns the maximum amount the taxi driver can earn. * * @param pickup an array of pickup locations (in km) * @param drop an array of drop-off locations (in km) * @param tip an array of tips offered by the riders * @return the maximum total earnings */ public static long taxiDriver(long[] pickup, long[] drop, int[] tip) { int n = pickup.length; // Create an array of rides. Ride[] rides = new Ride[n]; for (int i = 0; i < n; i++) { rides[i] = new Ride(pickup[i], drop[i], tip[i]); } // Sort the rides by drop-off location. Arrays.sort(rides, (a, b) -> Long.compare(a.drop, b.drop)); // Create a DP array where dp[i] will be the maximum earnings using the first i rides. long[] dp = new long[n + 1]; dp[0] = 0; // For binary search, store the drop times in a separate array. long[] endTimes = new long[n]; for (int i = 0; i < n; i++) { endTimes[i] = rides[i].drop; } // Process each ride. for (int i = 1; i <= n; i++) { // Current ride is rides[i-1] (because dp array is 1-indexed). Ride currentRide = rides[i - 1]; // Find the last ride that ends at or before currentRide.pickup. // We want the rightmost index j such that endTimes[j] <= currentRide.pickup. int j = Arrays.binarySearch(endTimes, 0, i - 1, currentRide.pickup); if (j < 0) { // If not found, binarySearch returns (-(insertion point) - 1). // The insertion point is the index of the first element greater than currentRide.pickup. j = -j - 2; } // Candidate earnings if we take the current ride: // dp[j+1] is the best earnings before current ride's pickup. long takeRide = (j >= 0 ? dp[j + 1] : 0) + currentRide.profit; // Option 1: Skip current ride; Option 2: Take current ride. dp[i] = Math.max(dp[i - 1], takeRide); } return dp[n]; } // Example usage public static void main(String[] args) { long[] pickup = {0, 2, 9, 10, 11, 12}; long[] drop = {5, 9, 11, 11, 14, 17}; int[] tip = {1, 2, 3, 2, 2, 1}; long maxEarnings = taxiDriver(pickup, drop, tip); System.out.println("Maximum earnings: " + maxEarnings); // Expected output: 20 } }

Explanation

  1. Ride Class:

    • The Ride class holds the pickup, drop, and computed profit for each ride.
  2. Sorting:

    • We sort the rides by their drop-off locations. This helps in easily finding the latest ride that does not conflict with the current one.
  3. Dynamic Programming Array:

    • dp[i] holds the maximum earnings possible by considering the first ii rides (after sorting).
    • For each ride (indexed by i1i-1 in the rides array), we use binary search on the endTimes array to find the ride with the largest drop-off that is less than or equal to the current ride’s pickup. Let that index be jj.
    • If we take the current ride, our total earnings become dp[j+1] + currentRide.profit.
    • Otherwise, we just carry over dp[i-1] (skipping the current ride).
    • We take the maximum of these two choices.
  4. Example:

    • With the example input, the DP process will eventually compute the maximum earning as 20, matching the provided explanation.

This solution meets the constraints and should work efficiently even when there are up to 10410^4 rides.

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