Find the first circular tour that visits all petrol pumps

Suppose there is a circle. There are n petrol pumps on that circle. You are given two sets of data.

1. The amount of petrol that every petrol pump has.
2. Distance from that petrol pump to the next petrol pump.

Calculate the first point from where a truck will be able to complete the circle (The truck will stop at each petrol pump and it has infinite capacity). Expected time complexity is O(n). Assume for 1 litre petrol, the truck can go 1 unit of distance.

For example, let there be 4 petrol pumps with amount of petrol and distance to next petrol pump value pairs as {4, 6}, {6, 5}, {7, 3} and {4, 5}. The first point from where truck can make a circular tour is 2nd petrol pump. Output should be “start = 1” (index of 2nd petrol pump).

A Simple Solution is to consider every petrol pumps as starting point and see if there is a possible tour. If we find a starting point with feasible solution, we return that starting point. The worst case time complexity of this solution is O(n^2).

We can use a Queue to store the current tour. We first enqueue first petrol pump to the queue, we keep enqueueing petrol pumps till we either complete the tour, or current amount of petrol becomes negative. If the amount becomes negative, then we keep dequeueing petrol pumps till the current amount becomes positive or queue becomes empty.

Instead of creating a separate queue, we use the given array itself as queue. We maintain two index variables start and end that represent rear and front of queue.

C/C++

 // C program to find circular tour for a truck #include    // A petrol pump has petrol and distance to next petrol pump struct petrolPump {   int petrol;   int distance; };    // The function returns starting point if there is a possible solution, // otherwise returns -1 int printTour(struct petrolPump arr[], int n) {     // Consider first petrol pump as a starting point     int start = 0;     int end =  1;        int curr_petrol = arr[start].petrol - arr[start].distance;        /* Run a loop while all petrol pumps are not visited.       And we have reached first petrol pump again with 0 or more petrol */     while (end != start || curr_petrol < 0)     {         // If curremt amount of petrol in truck becomes less than 0, then         // remove the starting petrol pump from tour         while (curr_petrol < 0 && start != end)         {             // Remove starting petrol pump. Change start             curr_petrol -= arr[start].petrol - arr[start].distance;             start = (start + 1)%n;                // If 0 is being considered as start again, then there is no             // possible solution             if (start == 0)                return -1;         }            // Add a petrol pump to current tour         curr_petrol += arr[end].petrol - arr[end].distance;            end = (end + 1)%n;     }        // Return starting point     return start; }    // Driver program to test above functions int main() {     struct petrolPump arr[] = {{6, 4}, {3, 6}, {7, 3}};        int n = sizeof(arr)/sizeof(arr);     int start = printTour(arr, n);        (start == -1)? printf("No solution"): printf("Start = %d", start);        return 0; }

Java

 //Java program to find circular tour for a truck    public class Petrol  {     // A petrol pump has petrol and distance to next petrol pump     static class petrolPump     {         int petrol;         int distance;                    // constructor         public petrolPump(int petrol, int distance)          {             this.petrol = petrol;             this.distance = distance;         }     }            // The function returns starting point if there is a possible solution,     // otherwise returns -1     static int printTour(petrolPump arr[], int n)     {           int start = 0;         int end = 1;         int curr_petrol = arr[start].petrol - arr[start].distance;                    // If current amount of petrol in truck becomes less than 0, then         // remove the starting petrol pump from tour         while(end != start || curr_petrol < 0)         {                            // If current amount of petrol in truck becomes less than 0, then             // remove the starting petrol pump from tour             while(curr_petrol < 0 && start != end)             {                 // Remove starting petrol pump. Change start                 curr_petrol -= arr[start].petrol - arr[start].distance;                 start = (start + 1) % n;                                    // If 0 is being considered as start again, then there is no                 // possible solution                 if(start == 0)                     return -1;             }             // Add a petrol pump to current tour             curr_petrol += arr[end].petrol - arr[end].distance;                            end = (end + 1)%n;         }                    // Return starting point         return start;     }            // Driver program to test above functions     public static void main(String[] args)      {                    petrolPump[] arr = {new petrolPump(6, 4),                             new petrolPump(3, 6),                             new petrolPump(7, 3)};                    int start = printTour(arr, arr.length);                    System.out.println(start == -1 ? "No Solution" : "Start = " + start);         }    } //This code is contributed by Sumit Ghosh

Python

 # Python program to find circular tour for a track     # A petrol pump has petrol and distance to next petrol pimp class PetrolPump:            # Constructor to create a new node     def __init__(self,petrol, distance):         self.petrol = petrol         self.distance = distance     # The funtion return starting point if there is a possible # solution otherwise returns -1  def printTour(arr):            n = len(arr)     # Consider first petrol pump as starting point     start = 0      end = 1             curr_petrol = arr[start].petrol - arr[start].distance             # Run a loop whie all petrol pumps are not visited     # And we have reached first petrol pump again with 0      # or more petrol     while(end != start or curr_petrol < 0 ):                    # If current amount of petrol pumps are not visited         # And we have reached first petrol pump again with         # 0 or more petrol          while(curr_petrol < 0 and start != end):                            # Remove starting petrol pump. Change start             curr_petrol -= arr[start].petrol - arr[start].distance             start = (start +1)%n                            # If 0 is being considered as start again, then             # there is no possible solution             if start == 0:                 return -1            # Add a petrol pump to current tour         curr_petrol += arr[end].petrol - arr[end].distance                     end = (end +1) % n        return start     # Driver program to test above function arr = [PetrolPump(6,4), PetrolPump(3,6), PetrolPump(7,3)] start = printTour(arr)    print "No solution" if start == -1 else "start =", start    # This code is contributed by Nikhil Kumar Singh(nickzuck_007)

C#

 // C# program to find circular  // tour for a truck  using System;    class GFG {     // A petrol pump has petrol and      // distance to next petrol pump      public class petrolPump     {         public int petrol;         public int distance;            // constructor          public petrolPump(int petrol,                            int distance)         {             this.petrol = petrol;             this.distance = distance;         }     }        // The function returns starting point      // if there is a possible solution,      // otherwise returns -1      public static int printTour(petrolPump[] arr,                                  int n)     {         int start = 0;         int end = 1;         int curr_petrol = arr[start].petrol -                            arr[start].distance;            // If current amount of petrol in           // truck becomes less than 0, then          // remove the starting petrol pump from tour          while (end != start || curr_petrol < 0)         {                // If current amount of petrol in             // truck becomes less than 0, then              // remove the starting petrol pump from tour              while (curr_petrol < 0 && start != end)             {                 // Remove starting petrol pump.                 // Change start                  curr_petrol -= arr[start].petrol -                                 arr[start].distance;                 start = (start + 1) % n;                    // If 0 is being considered as                  // start again, then there is no                  // possible solution                  if (start == 0)                 {                     return -1;                 }             }                            // Add a petrol pump to current tour              curr_petrol += arr[end].petrol -                             arr[end].distance;                end = (end + 1) % n;         }            // Return starting point          return start;     }        // Driver Code     public static void Main(string[] args)     {         petrolPump[] arr = new petrolPump[]         {             new petrolPump(6, 4),             new petrolPump(3, 6),             new petrolPump(7, 3)         };            int start = printTour(arr, arr.Length);            Console.WriteLine(start == -1 ? "No Solution" :                                     "Start = " + start);     } }    // This code is contributed by Shrikant13

Output:

start = 2

Time Complexity: Seems to be more than linear at first look. If we consider the items between start and end as part of a circular queue, we can observe that every item is enqueued at most two times to the queue. The total number of operations is proportional to total number of enqueue operations. Therefore the time complexity is O(n).

Auxiliary Space: O(1)