# Solution For Minimum Time To Complete Trips

Problem:

You are given n trips (numbered from 0 to n-1) and a 2D array trips where trips[i] = [startTime_i, endTime_i, passengersCount_i] indicates that the ith trip started at startTime_i and ended at endTime_i with passengersCount_i passengers.

You have a train that can travel on the s single track of length 1 unit. This train travels from any point to any other point on the track and it takes 1 unit of time to reach its destination. However, the train cannot pass other trips or stop during a trip.

Let’s say the train starts at the position pos and has an initial speed of speed units per second, which means that:

- The train moves 1 unit of distance to the right each second.
- If the train is at a position x different from pos, it takes |x – pos| * speed seconds to reach position x.

The train can choose its initial position, but once it starts moving, it cannot stop until it reaches the end of the track.

The train should find the minimum time needed to visit all the trips. You can assume that the starting speed of the train is such that it can visit all the trips.

Return the minimum time in seconds needed to visit all the trips.

Example 1:

Input: trips = [[9,12,5],[3,9,4],[4,6,5],[12,15,5],[7,11,3]], speed = 5, starting position = 2

Output: 34

Explanation: The best strategy is:

– Starting at position 2:

– Go to trip 1. Time needed: current position / speed = 2 / 5 = 0.4 seconds.

– Go to trip 2. Time needed: (9 – 2) / 5 = 1.4 seconds.

– Pick up all passengers. Time needed: 0 seconds.

– Go to trip 3. Time needed: (4 – 9) / 5 = 1 seconds.

– Drop off all passengers. Time needed: 0 seconds.

– Go to trip 0. Time needed: (6 – 4) / 5 = 0.4 seconds.

– Pick up all passengers. Time needed: 0 seconds.

– Go to trip 4. Time needed: (7 – 6) / 5 = 0.2 seconds.

– Pick up all passengers. Time needed: 0 seconds.

– Go to trip 3. Time needed: (12 – 7) / 5 = 1 seconds.

– Drop off all passengers. Time needed: 0 seconds.

– Go to trip 4. Time needed: (12 – 7) / 5 = 1 seconds.

– Drop off all passengers. Time needed: 0 seconds.

– Go to the end. Time needed: (15 – 12) / 5 = 0.6 seconds.

– Total time needed: 0.4 + 1.4 + 1 + 0.4 + 0.2 + 1 + 1 + 0.6 = 6 seconds.

Solution:

To solve the problem, we can use binary search to find the minimum time required to complete all the trips. We can assume that the time required is within the range [0, max_end_time], where max_end_time is the maximum end time among all the trips.

For each value of time in the range [0, max_end_time], we can simulate the train journey starting from each position on the single track.

For each starting position and speed, we can calculate the time required to complete all the trips, using dynamic programming.

The state of the dynamic programming is represented as a tuple (pos, speed, i), where pos is the current position of the train on the single track, speed is the current speed of the train, and i is the index of the next trip to visit.

The base case of the dynamic programming is when i = n, which means all the trips have been visited. The cost of this state is 0.

The transition of the dynamic programming is defined as follows:

For each trip j such that endTime_j <= time and i <= j, we can consider two options:

- If the train has not yet picked up any passengers at trip j, we can pick up all the passengers and move to the next trip. The cost of this option is the time required to move from pos to startTime_j, plus the time required to pick up all the passengers and move to the next trip, plus the cost of the next state (pos = endTime_j, speed = 1, i = j + 1).
- If the train has already picked up the passengers at trip j, we can move directly to the next trip. The cost of this option is the time required to move from pos to endTime_j, plus the cost of the next state (pos = endTime_j, speed = 1, i = j + 1).

The final answer is the minimum cost among all the starting positions and speeds.

The time complexity of this approach is O(n^3 log(max_end_time)), where n is the number of trips and max_end_time is the maximum end time among all the trips. The space complexity is O(n^2).

Code:

Here is the Python code for the solution:

class Solution:

def minTime(self, trips: List[List[int]], speed: int, start: int) -> int:

n = len(trips)

max_end_time = max(trips, key=lambda x: x[1])[1]

```
def dp(pos, speed, i, memo):
if i == n:
return 0
if (pos, speed, i) in memo:
return memo[(pos, speed, i)]
time = trips[i][0]
min_cost = float('inf')
for j in range(i, n):
if trips[j][1] <= time:
cost_1 = abs(trips[j][0] - pos) / speed + dp(trips[j][1], 1, j+1, memo)
cost_2 = (trips[j][1] - pos) / speed + dp(trips[j][1], 1, j+1, memo)
min_cost = min(min_cost, cost_1, cost_2)
else:
break
memo[(pos, speed, i)] = min_cost
return min_cost
left, right = 0, max_end_time
while left < right:
mid = (left + right) // 2
if dp(start, speed, 0, {}) <= mid:
right = mid
else:
left = mid + 1
return left
```

## Step by Step Implementation For Minimum Time To Complete Trips

class Solution { public int minTime(int n, int[][] edges, int[][] edgesTo) { // This problem can be solved using a breadth-first search algorithm. // We start by initializing a visited array to keep track of which nodes have been visited. // We also initialize a queue to keep track of the nodes that we need to visit. // We start by enqueuing the first node in the queue and marking it as visited. // Then, we iterate through the queue until it is empty. // For each node in the queue, we iterate through its edges and add the nodes that have not been visited yet to the queue. // We also update the visited array for these nodes. // Finally, we return the length of the visited array, which will be the minimum number of nodes that need to be visited. boolean[] visited = new boolean[n]; Queuequeue = new LinkedList<>(); queue.add(0); visited[0] = true; while (!queue.isEmpty()) { int node = queue.poll(); for (int i = 0; i < edges[node].length; i++) { int nextNode = edges[node][i]; if (!visited[nextNode]) { queue.add(nextNode); visited[nextNode] = true; } } } return visited.length; } }

def minimumTimeToCompleteTrips(trips, capacity): # sort the trips in increasing order of start time trips.sort(key = lambda x: x[1]) # initialize the current number of passengers and the result num_passengers = 0 result = 0 # go through all the trips for i in range(len(trips)): # if there are not enough passengers to complete the current trip, # we need to wait until we have enough passengers while num_passengers < trips[i][0]: num_passengers += 1 result += 1 # once we have enough passengers, we can complete the trip num_passengers -= trips[i][0] # return the result return result

var minimumTimeToCompleteTrips = function(tripTimes, numTrips, numStops) { // your code here };

int findMinCompletionTime(vector& trips, int capacity) { // check for invalid inputs if (trips.size() == 0 || capacity <= 0) { return -1; } // sort the trips by their starting time sort(trips.begin(), trips.end(), [](const vector & a, const vector & b) { return a[1] < b[1]; }); // initialize a priority queue to store the end times of the trips // the trips will be stored in reverse order of their end times // so that we can always pop off the trip that ends first priority_queue , greater > pq; int currTime = 0; int i = 0; // while there are still trips remaining or there are still people // on the bus while (i < trips.size() || !pq.empty()) { // if there are still trips remaining if (i < trips.size()) { // if the current trip starts after the last person has left the bus if (trips[i][1] >= currTime) { // add the trip to the priority queue pq.push(trips[i][2]); // move on to the next trip i++; } else { // if the current trip starts before the last person has left the bus // if there is still room on the bus if (pq.size() < capacity) { // add the trip to the priority queue pq.push(trips[i][2]); // move on to the next trip i++; } else { // if there is no room on the bus // pop off the trip that ends first pq.pop(); } } } else { // if there are no more trips remaining // pop off the trip that ends first pq.pop(); } // update the current time currTime = pq.top(); } return currTime; }

using System; public class Solution { // Function to calculate the minimum time to complete // the given number of trips static int minTime(int numTrips, int[,] trips) { // Sort the trips according to their starting time Array.Sort(trips, new Comparison((a, b) => a[1].CompareTo(b[1]))); // Variable to store the current end time int end = 0; // Variable to store the result int result = 0; // Iterate over all the trips for (int i = 0; i < numTrips; i++) { // If the current trip starts before the end // of the previous trip, then update the end time if (trips[i][1] < end) end = Math.Max(end, trips[i][2]); // Else, update the result else { result += end - trips[i][1]; end = trips[i][2]; } } // Return the result return result; } // Driver code public static void Main() { int numTrips = 4; int[,] trips = {{2, 1, 5}, {3, 3, 7}, {5, 2, 8}, {6, 1, 9}}; Console.WriteLine(minTime(numTrips, trips)); } }