# 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.

- The amount of petrol that every petrol pump has.
- 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 the 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 a starting point and see if there is a possible tour. If we find a starting point with a feasible solution, we return that starting point. The worst case time complexity of this solution is O(n^2).

An efficient approach is to **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 the current amount of petrol becomes negative. If the amount becomes negative, then we keep dequeuing petrol pumps until the queue becomes empty.

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

Below image is a dry run of the above approach:

Below is the implementation of the above approach:

## C++

`// C++ program to find circular tour for a truck` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// A petrol pump has petrol and distance to next petrol pump` `class` `petrolPump` `{` ` ` `public` `:` ` ` `int` `petrol;` ` ` `int` `distance;` `};` `// The function returns starting point if there is a possible solution,` `// otherwise returns -1` `int` `printTour(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 code` `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;` `}` `// This code is contributed by rathbhupendra` |

## 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 truck` `# In this approach we will start the tour from the first petrol pump` `# then while moving to the next pumps in the loop we will store the cumulative` `# information that whether we have a deficit of petrol at the current pump or not` `# If there is a deficit then we will add it to the deficit value calculated` `# till the previous petrol pump and then update the starting point to the next pump` `# and reset the petrol available in the truck as 0` `# This function return starting point if there is a possible` `# solution otherwise returns -1` `def` `printTour(arr,n):` ` ` ` ` `# Consider first petrol pump as starting point` ` ` `start ` `=` `0` ` ` `# These two variable will keep tracking if there is` ` ` `# surplus(s) or deficit(d) of petrol in the truck` ` ` `s ` `=` `0` `# petrol available the truck till now` ` ` `d ` `=` `0` `# deficit of petrol till visiting this petrol pump` ` ` ` ` `# Start from the first petrol pump and complete one loop` ` ` `# of visiting all the petrol pumps and keep updating s and d at each pump` ` ` `for` `i ` `in` `range` `(n):` ` ` `s ` `+` `=` `arr[i][` `0` `] ` `-` `arr[i][` `1` `]` ` ` `if` `s < ` `0` `: ` `# the truck has a deficit of petrol` ` ` `start ` `=` `i` `+` `1` `# change the starting point` ` ` `d ` `+` `=` `s ` `# storing the deficit of petrol till current petrol pump` ` ` `s ` `=` `0` `# starting again from new station` ` ` ` ` `# when we reach first petrol pump again and sum of the petrol available at the truck` ` ` `# and the petrol deficit till now is 0 or more petrol then return the starting point` ` ` `# else return -1` ` ` `return` `start ` `if` `(s` `+` `d)>` `=` `0` `else` `-` `1` ` ` ` ` `# Driver program to test above function` `arr ` `=` `[[` `6` `,` `4` `], [` `3` `,` `6` `], [` `7` `,` `3` `]]` `start ` `=` `printTour(arr,` `3` `)` `if` `start ` `=` `=` `-` `1` `:` ` ` `print` `(` `"No Solution Possible !!!"` `)` `else` `:` ` ` `print` `(` `"start = {}"` `.` `format` `(start))` `# This code is contributed by Antara Das(anny)` |

## 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` |

## Javascript

`<script>` ` ` `// JavaScript program to find circular tour for a truck` ` ` `// A petrol pump has petrol and distance to next petrol pump` ` ` `class petrolPump {` ` ` `constructor(petrol, distance) {` ` ` `this` `.petrol = petrol;` ` ` `this` `.distance = distance;` ` ` `}` ` ` `};` ` ` `// The function returns starting point if there is a possible solution,` ` ` `// otherwise returns -1` ` ` `const printTour = (arr, n) => {` ` ` `// Consider first petrol pump as a starting point` ` ` `let start = 0;` ` ` `let end = 1;` ` ` `let 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 code` ` ` `let arr = [` `new` `petrolPump(6, 4), ` `new` `petrolPump(3, 6), ` `new` `petrolPump(7, 3)];` ` ` `let n = arr.length;` ` ` `let start = printTour(arr, n);` ` ` `(start == -1) ? document.write(` `"No solution"` `) : document.write(`Start = ${start}`);` `// This code is contributed by rakeshsahni` `</script>` |

**Output:**

start = 2

**Time Complexity:** We are visiting each petrol pump exactly once, therefore the time complexity is **O(n)**

**Auxiliary Space: **O(1)

Another efficient solution can be to find out the first petrol pump where the amount of petrol is greater than or equal to the distance to be covered to reach the next petrol pump. Now we mark that petrol pump as **start **and now we check whether we can finish the journey towards the end point. If in the middle, at any petrol pump, the amount of petrol is less than the distance to be covered to reach the next petrol pump, then we can say we cannot complete the circular tour from **start**. We again try to find out the next point from where we can start our journey i.e. the next petrol pump where the amount of petrol is greater than or equal to the distance to be covered and we mark it as **start**. We need not look at any petrol pump in between the initial petrol pump marked as **start **and the new **start **as we know that we cannot complete the journey if we start from any middle petrol pump because eventually we will arrive at a point where amount of petrol is less than the distance. Now we repeat the process until we reach the last petrol pump and update our **start **as and when required. After we reach our last petrol pump, we try to reach our first petrol pump from the last and let’s say we have a remaining amount of petrol as **curr_petrol**. Now we again start traveling from the first petrol pump and take the advantage of our **curr_petrol **and try to reach the **start**. If we can reach the **start**, then we may conclude that **start **can be our starting point.

Below is the implementation of the above approach:

## C++

`// C++ program to find circular tour for a truck` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// A petrol pump has petrol and distance to next petrol pump` `class` `petrolPump {` `public` `:` ` ` `int` `petrol;` ` ` `int` `distance;` `};` `// The function returns starting point if there is a` `// possible solution, otherwise returns -1` `int` `printTour(petrolPump arr[], ` `int` `n)` `{` ` ` `int` `start;` ` ` `for` `(` `int` `i = 0; i < n; i++) {` ` ` `// Identify the first petrol pump from where we` ` ` `// might get a full circular tour` ` ` `if` `(arr[i].petrol >= arr[i].distance) {` ` ` `start = i;` ` ` `break` `;` ` ` `}` ` ` `}` ` ` `// To store the excess petrol` ` ` `int` `curr_petrol = 0;` ` ` `int` `i;` ` ` `for` `(i = start; i < n;) {` ` ` `curr_petrol += (arr[i].petrol - arr[i].distance);` ` ` `// If at any point remaining petrol is less than 0,` ` ` `// it means that we cannot start our journey from` ` ` `// current start` ` ` `if` `(curr_petrol < 0) {` ` ` `// We move to the next petrol pump` ` ` `i++;` ` ` `// We try to identify the next petrol pump from` ` ` `// where we might get a full circular tour` ` ` `for` `(; i < n; i++) {` ` ` `if` `(arr[i].petrol >= arr[i].distance) {` ` ` `start = i;` ` ` `// Reset rem_petrol` ` ` `curr_petrol = 0;` ` ` `break` `;` ` ` `}` ` ` `}` ` ` `}` ` ` `else` `{` ` ` `// Move to the next petrolpump if curr_petrol is` ` ` `// >= 0` ` ` `i++;` ` ` `}` ` ` `}` ` ` `// If remaining petrol is less than 0 while we reach the` ` ` `// first petrol pump, it means no circular tour is` ` ` `// possible` ` ` `if` `(curr_petrol < 0) {` ` ` `return` `-1;` ` ` `}` ` ` `for` `(` `int` `j = 0; j < start; j++) {` ` ` `curr_petrol += (arr[j].petrol - arr[j].distance);` ` ` `// If remaining petrol is less than 0 at any point` ` ` `// before we reach initial start, it means no` ` ` `// circular tour is possible` ` ` `if` `(curr_petrol < 0) {` ` ` `return` `-1;` ` ` `}` ` ` `}` ` ` `// If we have successfully reached intial_start, it` ` ` `// means can get a circular tour from final_start, hence` ` ` `// return it` ` ` `return` `start;` `}` `// Driver code` `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;` `}` `// This code is contributed by supratik_mitra` |

## Javascript

`<script>` ` ` `// JavaScript program to find circular tour for a truck` ` ` `// A petrol pump has petrol and distance to next petrol pump` ` ` `class petrolPump {` ` ` `constructor(petrol, distance) {` ` ` `this` `.petrol = petrol;` ` ` `this` `.distance = distance;` ` ` `}` ` ` `};` ` ` `// The function returns starting point if there is a` ` ` `// possible solution, otherwise returns -1` ` ` `const printTour = (arr, n) => {` ` ` `let start;` ` ` `for` `(let i = 0; i < n; i++)` ` ` `{` ` ` ` ` `// Identify the first petrol pump from where we` ` ` `// might get a full circular tour` ` ` `if` `(arr[i].petrol >= arr[i].distance) {` ` ` `start = i;` ` ` `break` `;` ` ` `}` ` ` `}` ` ` `// To store the excess petrol` ` ` `let curr_petrol = 0;` ` ` `let i;` ` ` `for` `(i = start; i < n;)` ` ` `{` ` ` `curr_petrol += (arr[i].petrol - arr[i].distance);` ` ` `// If at any point remaining petrol is less than 0,` ` ` `// it means that we cannot start our journey from` ` ` `// current start` ` ` `if` `(curr_petrol < 0) {` ` ` `// We move to the next petrol pump` ` ` `i++;` ` ` `// We try to identify the next petrol pump from` ` ` `// where we might get a full circular tour` ` ` `for` `(; i < n; i++) {` ` ` `if` `(arr[i].petrol >= arr[i].distance) {` ` ` `start = i;` ` ` `// Reset rem_petrol` ` ` `curr_petrol = 0;` ` ` `break` `;` ` ` `}` ` ` `}` ` ` `}` ` ` `else` `{` ` ` `// Move to the next petrolpump if curr_petrol is` ` ` `// >= 0` ` ` `i++;` ` ` `}` ` ` `}` ` ` `// If remaining petrol is less than 0 while we reach the` ` ` `// first petrol pump, it means no circular tour is` ` ` `// possible` ` ` `if` `(curr_petrol < 0) {` ` ` `return` `-1;` ` ` `}` ` ` `for` `(let j = 0; j < start; j++) {` ` ` `curr_petrol += (arr[j].petrol - arr[j].distance);` ` ` `// If remaining petrol is less than 0 at any point` ` ` `// before we reach initial start, it means no` ` ` `// circular tour is possible` ` ` `if` `(curr_petrol < 0) {` ` ` `return` `-1;` ` ` `}` ` ` `}` ` ` `// If we have successfully reached intial_start, it` ` ` `// means can get a circular tour from final_start, hence` ` ` `// return it` ` ` `return` `start;` ` ` `}` ` ` ` ` `// Driver code` ` ` `let arr = [` `new` `petrolPump(6, 4), ` `new` `petrolPump(3, 6), ` `new` `petrolPump(7, 3)];` ` ` `let n = arr.length;` ` ` `let start = printTour(arr, n);` ` ` `(start == -1) ? document.write(` `"No solution"` `) : document.write(`Start = ${start}`);` ` ` `// This code is contributed by rakeshsahni` `</script>` |

**Output:**

start = 2

**Time Complexity:** O(n)

**Auxiliary Space:** O(1)

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