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 <stdio.h> // 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[0]); 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)

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)

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