Find the first circular tour that visits all petrol pumps

There is a circle, you are given 2 values->

  • The amount of petrol in every petrol pump
  • Distance of petrol pump to the next petrol pump
Calculate the first point from where a truck will be able to complete the circle(truck stops at each pump and has infinite capacity). Assume for 1ltr petrol, the truck can go 1 unit of distance.

Example- there are 4 petrol pumps with amount of petrol and distance to next petrol pump in pairs as {4,3},{6,5},{7,3},{4,5}. The first point from where the truck can make a circular tour is 2nd petrol pump. 

Output should be "start=1" (index of 2nd petrol pump)

An easy solution is to consider ever petrol pump and check whether it can create a circular tour, we use queue to store the values. We first will enqueue first petrol pump to the queue, we keep enqueuing petrol pumps until petrol amount becomes negative, then we dequeue petrol pumps until queue becomes empty.

Code-->

class petrolpump{

    public:

    int petrol;

    int distance;

};

//function returns starting point else returns -1

int tour(petrolpump arr[], int n)

{

    int start=0;  //first petrolpump is the starting point

    int end=1;

    int current_petrol= arr[start].petrol - arr[start].distance;

//run loop to visit all petrol pumps now

while(end!=start || current_petrol<0)

{        //if current amount of petrol becomes 0 or negative then remove starting petrol pump from tour

        while(current_petrol<0 && start!=end)

        {    current_petrol-=arr[start].petrol -arr[start].distance;

              start= (start+1)% n;

              //if 0 is being considered as start again, then theres no solution

               if(start==0)

                    return -1;

        }

    current_petrol+=arr[end].petrol- arr[end].distance;

    end=(end+1)%n;

}

return start;

}

int main()

{

    petrolPump arr[] = {{6, 4}, {3, 6}, {7, 3}};

    int n = sizeof(arr)/sizeof(arr[0]);

    int start = printTour(arr, n);

    (start == -1)? cout<<"No solution": cout<<"Start = "<<start;

    return 0;

}


                

Popular posts from this blog

Finding the Subarrays (Hackerearth)

Image
Finding subarrays can become a cumbersome task to handle if we don't know the right approach for it. In this particular question we will be using a new method.  Pair is used to combine together two values which may be different in type. Pair provides a way to store two heterogeneous objects as a single unit. Such as vector<first attribute, second attribute>. To access these attributes we use keywords 'first' and 'second' Approach - Input values and find total sum of all terms Create a loop and calculate average term by term, use logic ->  sum= first element this_average=sum/elements in the sum SZ= Elements left in array= size-elements in sum other_average=(total-sum)/SZ Use if loop to find whether this_average is bigger than other_average or not. If true, input the (i+1)th and (j+1)th element in the pair, Increment counter by 1 Sort the pair and output it. Code- #include< bits/stdc++.h > using namespace std ; int main () {       ...
Read more

Palindrome Index (Hackerrank)

Image
Here we have another Hackerrank question on String manipulation, this one involving the classic Palindrome index.  Question-  Approach - Pretty simple, we just need to check values one by one, iterating a loop from index 0 being compared to index n, where n is the last value in the index, narrowing it down to inner values one by one. The point where the elements do not stay equal is the point where element needs to be removed in order to make the string into a palindrome  The steps are -  initialise i = 0; j= string.length()-1 //last value in the string Run a loop from i to middle of the string, i increments while j decrements, bringing us a section to focus on within the string Run an if condition which checks whether character at i is equal to j or not If it is not equal, we check if the element after character at i is equal to j or not If yes, we check whether the element before j is equal to character at i and character after i is subsequently equal to element at...
Read more

Sherlock and Array (Hackerrank)

Image
Sometimes it's better to stop and first formulate a way to solve a problem before jumping right into it. Why? Well because it gives you more time to better understand a problem and secondly, in some cases even a simpler way to solve a problem.  Remember kids! When solving algorithms, Math is your friend! I realized that a little late into this question and when i did, it was just a 5 min snippet. Question-  Approach - It's quite simple and straight forward.  Instead of dividing array into two halves, getting their sums, checking whether the sums are equal and trying to find out if an element which causes sum to be unequal exist or not, here's a better approach. array - [5,6,8,11] , here we have 5+6=11 in LHS and 11 in RHS with 8 being an extra element. Imagine LHS and RHS as same, ex - x+y+x= sum. i.e 2x= sum-y. Here we just need to check if this equality holds, if it does we print YES and add value to x, else we print NO Code- 
Read more