There are n gas stations along a circular tour, where the amount of gas at the ith gas station is gas[i]. You have a car with a gas tank of unlimited capacity and it costs cost[i] of gas to travel from the ith station to its next station. You begin the journey with an empty tank at one of the gas station.
Given two integer arrays gas[] and cost[], the task is to return the starting gas station's index if you want to travel around the circular tour once in the clockwise direction, otherwise return -1.
Note: If a solution exists, it is guaranteed to be unique.
Examples:
Input: gas[] = [4, 5, 7, 4], cost[] = [6, 6, 3, 5]
Output: 2
Explanation: Start at gas station at index 2 and fill up with 7 units of gas. Your tank = 0 + 7 = 7
- Travel to station 3. Available gas = (7 - 3 + 4) = 8.
- Travel to station 0. Available gas = (8 - 5 + 4) = 7.
- Travel to station 1. Available gas = (7 - 6 + 5) = 6.
- Return to station 2. Available gas = (6 - 6) = 0.
Therefore, return 2 as the starting index.
Input: gas[] = [1, 2 ,3 ,4, 5], cost[] = [3, 4, 5, 1, 2]
Output: 3
Explanation: Start at gas station 3 (index 3) and fill up with 4 units of gas. Your tank = 0 + 4 = 4
- Travel to station 4. Available gas = 4 - 1 + 5 = 8
- Travel to station 0. Available gas = 8 - 2 + 1 = 7
- Travel to station 1. Available gas= 7 - 3 + 2 = 6
- Travel to station 2. Available gas = 6 - 4 + 3 = 5
- Travel to station 3. The cost is 5. The gas is just enough to travel back to station 3.
Therefore, return 3 as the starting index.
Input: arr[] = [3, 9], cost[] = [7, 6]
Output: -1
Explanation: There is no gas station to start with such that you can complete the tour.
[Naive Approach] Considering Every Index as Starting Point - O(n^2) Time and O(1) Space
The simplest approach is to consider each index as a starting point and check if a car can complete the circular tour starting from that index. If we find a valid starting point, we will return it.
C++ // C++ program to find starting index of circular Tour // by considering each index as starting point. #include <iostream> #include <vector> using namespace std; int startStation(vector<int> &gas, vector<int> &cost) { int n = gas.size(); int startIdx = -1; for(int i = 0; i < n; i++) { // Initially car tank is empty int currGas = 0; bool flag = true; for (int j = 0; j < n; j++){ // Circular Index int idx = (i + j) % n; currGas = currGas + gas[idx] - cost[idx]; // If currGas is less than zero, then it isn't // possible to proceed further with this starting point if(currGas < 0) { flag = false; break; } } // If flag is true, then we have found // the valid starting point if(flag) { startIdx = i; break; } } return startIdx; } int main() { vector<int> gas = {1, 2, 3, 4, 5}; vector<int> cost = {3, 4, 5, 1, 2}; cout << startStation(gas, cost) << endl; return 0; } // C program to find starting index of circular Tour // by considering each index as starting point. #include <stdio.h> int startStation(int gas[], int cost[], int n) { int startIdx = -1; for (int i = 0; i < n; i++) { // Current Available gas int currGas = 0; int flag = 1; for (int j = 0; j < n; j++) { // Circular Index int idx = (i + j) % n; currGas = currGas + gas[idx] - cost[idx]; // If Available gas is less than zero, then it isn't // possible to proceed further with this starting point if (currGas < 0) { flag = 0; break; } } // If flag is true, then we have found // the valid starting point if (flag) { startIdx = i; break; } } return startIdx; } int main() { int gas[] = {1, 2, 3, 4, 5}; int cost[] = {3, 4, 5, 1, 2}; int n = sizeof(gas) / sizeof(gas[0]); printf("%d\n", startStation(gas, cost, n)); return 0; } // Java program to find starting index of circular Tour // by considering each index as starting point. import java.util.*; class GfG { static int startStation(int[] gas, int[] cost) { int n = gas.length; int startIdx = -1; for (int i = 0; i < n; i++) { // Current Available gas int currGas = 0; boolean flag = true; for (int j = 0; j < n; j++) { // Circular Index int idx = (i + j) % n; currGas = currGas + gas[idx] - cost[idx]; // If Available gas is less than zero, then it isn't // possible to proceed further with this starting point if (currGas < 0) { flag = false; break; } } // If flag is true, then we have found // the valid starting point if (flag) { startIdx = i; break; } } return startIdx; } public static void main(String[] args) { int[] gas = {1, 2, 3, 4, 5}; int[] cost = {3, 4, 5, 1, 2}; System.out.println(startStation(gas, cost)); } } # Python program to find starting index of circular Tour # by considering each index as starting point. def startStation(gas, cost): n = len(gas) startIdx = -1 for i in range(n): # Current Available gas currGas = 0 flag = True for j in range(n): # Circular Index idx = (i + j) % n currGas += gas[idx] - cost[idx] # If Available gas is less than zero, then it isn't # possible to proceed further with this starting point if currGas < 0: flag = False break # If flag is true, then we have found # the valid starting point if flag: startIdx = i break return startIdx if __name__ == "__main__": gas = [1, 2, 3, 4, 5] cost = [3, 4, 5, 1, 2] print(startStation(gas, cost))
// C# program to find starting index of circular Tour // by considering each index as starting point. using System; class GfG { static int startStation(int[] gas, int[] cost) { int n = gas.Length; int startIdx = -1; for (int i = 0; i < n; i++) { // Current Available gas int currGas = 0; bool flag = true; for (int j = 0; j < n; j++) { // Circular Index int idx = (i + j) % n; currGas = currGas + gas[idx] - cost[idx]; // If Available gas is less than zero, then it isn't // possible to proceed further with this starting point if (currGas < 0) { flag = false; break; } } // If flag is true, then we have found // the valid starting point if (flag) { startIdx = i; break; } } return startIdx; } static void Main() { int[] gas = { 1, 2, 3, 4, 5 }; int[] cost = { 3, 4, 5, 1, 2 }; Console.WriteLine(startStation(gas, cost)); } } // JavaScript program to find starting index of circular Tour // by considering each index as starting point. function startStation(gas, cost) { let n = gas.length; let startIdx = -1; for (let i = 0; i < n; i++) { // Current Available gas let currGas = 0; let flag = true; for (let j = 0; j < n; j++) { // Circular Index let idx = (i + j) % n; currGas += gas[idx] - cost[idx]; // If Available gas is less than zero, then it isn't // possible to proceed further with this starting point if (currGas < 0) { flag = false; break; } } // If flag is true, then we have found // the valid starting point if (flag) { startIdx = i; break; } } return startIdx; } // Driver Code let gas = [1, 2, 3, 4, 5]; let cost = [3, 4, 5, 1, 2]; console.log(startStation(gas, cost)); [Expected Approach 1] Greedy Approach - O(n) Time and O(1) Space
We start by assuming the 0th index as the starting point for the circular tour. As we traverse the array, we calculate the available gas at each gas station, which is the previously available gas + gas[i] - cost[i]. If, at any station i, the available gas drops below zero, it indicates that a car cannot proceed to the next station (i + 1) from the current starting point. In such a case, we update the starting point to i + 1 and continue the process. After completing the traversal of the array, we check whether the starting point is valid for the circular tour.
If a car starts at gas station A and cannot reach gas station B, then any gas station located between A and B cannot help us reach B either. But why?
If we start at A and are unable to reach B, but we can reach all the stations up to B-1. Let's assume a gas station C (C<=B-1) located between station A and B. When we arrive at C from A, we must have had a positive amount of gas in our tank. Therefore, if we can't reach B starting with positive amount of gas at C, it would be impossible to reach B from C with a zero amount of gas.
C++ // C++ program to find starting index of circular Tour // using greedy approach #include <iostream> #include <vector> using namespace std; int startStation(vector<int> &gas, vector<int> &cost) { int n = gas.size(); int startIdx = 0; // Initially car tank is empty int currGas = 0; for(int i = 0; i < n; i++) { currGas = currGas + gas[i] - cost[i]; // If currGas becomes less than zero, then // It's not possible to proceed with this startIdx if(currGas < 0) { startIdx = i + 1; currGas = 0; } } // Checking if startIdx can be a valid // starting point for the Circular tour currGas = 0; for(int i = 0; i < n; i++) { // Circular Index int idx = (i + startIdx) % n; currGas = currGas + gas[idx] - cost[idx]; if(currGas < 0) return -1; } return startIdx; } int main() { vector<int> gas = {1, 2, 3, 4, 5}; vector<int> cost = {3, 4, 5, 1, 2}; cout << startStation(gas, cost) << endl; return 0; } // C program to find starting index of circular Tour // using greedy approach #include <stdio.h> int startStation(int gas[], int cost[], int n) { int startIdx = 0; // Initially car tank is empty int currGas = 0; for (int i = 0; i < n; i++) { currGas = currGas + gas[i] - cost[i]; // If currGas becomes less than zero, then // It's not possible to proceed with this startIdx if (currGas < 0) { startIdx = i + 1; currGas = 0; } } // Checking if startIdx can be a valid // starting point for the Circular tour. currGas = 0; for (int i = 0; i < n; i++) { // Circular Index int idx = (i + startIdx) % n; currGas = currGas + gas[idx] - cost[idx]; if (currGas < 0) return -1; } return startIdx; } int main() { int gas[] = {1, 2, 3, 4, 5}; int cost[] = {3, 4, 5, 1, 2}; int n = sizeof(gas) / sizeof(gas[0]); printf("%d\n", startStation(gas, cost, n)); return 0; } // Java program to find starting index of circular Tour // using greedy approach import java.util.*; class GfG { static int startStation(int[] gas, int[] cost) { int n = gas.length; int startIdx = 0; // Initially car tank is empty int currGas = 0; for (int i = 0; i < n; i++) { currGas = currGas + gas[i] - cost[i]; // If currGas becomes less than zero, then // It's not possible to proceed with this startIdx if (currGas < 0) { startIdx = i + 1; currGas = 0; } } // Checking if startIdx can be a valid // starting point for the Circular tour currGas = 0; for (int i = 0; i < n; i++) { // Circular Index int idx = (i + startIdx) % n; currGas = currGas + gas[idx] - cost[idx]; if (currGas < 0) return -1; } return startIdx; } public static void main(String[] args) { int[] gas = {1, 2, 3, 4, 5}; int[] cost = {3, 4, 5, 1, 2}; System.out.println(startStation(gas, cost)); } } # Python program to find starting index of circular Tour # using greedy approach def startStation(gas, cost): n = len(gas) startIdx = 0 # Initially car tank is empty currGas = 0 for i in range(n): currGas = currGas + gas[i] - cost[i] # If currGas becomes less than zero, then # It's not possible to proceed with this startIdx if currGas < 0: startIdx = i + 1 currGas = 0 # Checking if startIdx can be a valid # starting point for the Circular tour currGas = 0 for i in range(n): # Circular Index idx = (i + startIdx) % n currGas = currGas + gas[idx] - cost[idx] if currGas < 0: return -1 return startIdx if __name__ == "__main__": gas = [1, 2, 3, 4, 5] cost = [3, 4, 5, 1, 2] print(startStation(gas, cost))
// C# program to find starting index of a circular Tour // using greedy approach using System; class GfG { static int startStation(int[] gas, int[] cost) { int n = gas.Length; int startIdx = 0; // Initially car tank is empty int currGas = 0; for (int i = 0; i < n; i++) { currGas = currGas + gas[i] - cost[i]; // If currGas becomes less than zero, then // It's not possible to proceed with this startIdx if (currGas < 0) { startIdx = i + 1; currGas = 0; } } // Check if startIdx can be a valid // starting point for the Circular tour currGas = 0; for (int i = 0; i < n; i++) { // Circular Index int idx = (i + startIdx) % n; currGas = currGas + gas[idx] - cost[idx]; if (currGas < 0) return -1; } return startIdx; } static void Main() { int[] gas = { 1, 2, 3, 4, 5 }; int[] cost = { 3, 4, 5, 1, 2 }; Console.WriteLine(startStation(gas, cost)); } } // JavaScript program to find starting index of circular Tour // using greedy approach function startStation(gas, cost) { let n = gas.length; let startIdx = 0; // Initially car tank is empty let currGas = 0; for (let i = 0; i < n; i++) { currGas = currGas + gas[i] - cost[i]; // If currGas becomes less than zero, then // It's not possible to proceed with this startIdx if (currGas < 0) { startIdx = i + 1; currGas = 0; } } // Checking if startIdx can be a valid // starting point for the Circular tour currGas = 0; for (let i = 0; i < n; i++) { // Circular Index let idx = (i + startIdx) % n; currGas = currGas + gas[idx] - cost[idx]; if (currGas < 0) return -1; } return startIdx; } // driver code let gas = [1, 2, 3, 4, 5]; let cost = [3, 4, 5, 1, 2]; console.log(startStation(gas, cost));
[Expected Approach 2] Greedy Approach in One Pass - O(n) Time and O(1) Space
This approach is optimization for the previous one. After completing the entire traversal of the array, instead of checking the validity by circularly traversing from the starting index, we calculate the total gas remaining (net gas and the cost difference). If the difference is greater than or equal to zero, then it's obvious that the starting point is valid; otherwise, it is not possible to complete a circular loop.
C++ // C++ program to find starting index of circular Tour // using greedy approach in one pass #include <iostream> #include <vector> using namespace std; int startStation(vector<int> &gas, vector<int> &cost) { int n = gas.size(); // Variables to track total and current remaining gas int totalGas = 0; int currGas = 0; int startIdx = 0; // Traverse through each station to calculate remaining // gas in the tank, and total gas for(int i = 0; i < n; i++) { currGas += gas[i] - cost[i]; totalGas += gas[i] - cost[i]; // If currGas is negative, circular tour can't // start with this index, so update it to next one if(currGas < 0) { currGas = 0; startIdx = i + 1; } } // No solution exist if(totalGas < 0) return -1; return startIdx; } int main() { vector<int> gas = {1, 2, 3, 4, 5}; vector<int> cost = {3, 4, 5, 1, 2}; cout << startStation(gas, cost); return 0; } // C program to find starting index of circular Tour // using greedy approach in one pass #include <stdio.h> int startStation(int gas[], int cost[], int n) { // Variables to track total and current remaining gas int totalGas = 0; int currGas = 0; int startIdx = 0; // Traverse through each station to calculate remaining // gas in the tank, and total gas for (int i = 0; i < n; i++) { currGas += gas[i] - cost[i]; totalGas += gas[i] - cost[i]; // If currGas is negative, circular tour can't // start with this index, so update it to next one if (currGas < 0) { currGas = 0; startIdx = i + 1; } } // No solution exists if (totalGas < 0) return -1; return startIdx; } int main() { int gas[] = {1, 2, 3, 4, 5}; int cost[] = {3, 4, 5, 1, 2}; int n = sizeof(gas) / sizeof(gas[0]); printf("%d\n", startStation(gas, cost, n)); return 0; } // Java program to find starting index of circular Tour // using greedy approach in one pass class GfG { static int startStation(int[] gas, int[] cost) { int n = gas.length; // Variables to track total and current remaining gas int totalGas = 0; int currGas = 0; int startIdx = 0; // Traverse through each station to calculate remaining // gas in the tank, and total gas for (int i = 0; i < n; i++) { currGas += gas[i] - cost[i]; totalGas += gas[i] - cost[i]; // If currGas is negative, circular tour can't // start with this index, so update it to next one if (currGas < 0) { currGas = 0; startIdx = i + 1; } } // No solution exists if (totalGas < 0) return -1; return startIdx; } public static void main(String[] args) { int[] gas = { 1, 2, 3, 4, 5 }; int[] cost = { 3, 4, 5, 1, 2 }; System.out.println(startStation(gas, cost)); } } # Python program to find starting index of circular Tour # using greedy approach in one pass def startStation(gas, cost): n = len(gas) # Variables to track total and current remaining gas totalGas = 0 currGas = 0 startIdx = 0 # Traverse through each station to calculate remaining # gas in the tank, and total gas for i in range(n): currGas += gas[i] - cost[i] totalGas += gas[i] - cost[i] # If currGas is negative, circular tour can't # start with this index, so update it to next one if currGas < 0: currGas = 0 startIdx = i + 1 # No solution exists if totalGas < 0: return -1 return startIdx if __name__ == "__main__": gas = [1, 2, 3, 4, 5] cost = [3, 4, 5, 1, 2] print(startStation(gas, cost))
// C# program to find starting index of circular Tour // using greedy approach in one pass using System; class GfG { static int startStation(int[] gas, int[] cost) { int n = gas.Length; // Variables to track total and current remaining gas int totalGas = 0; int currGas = 0; int startIdx = 0; // Traverse through each station to calculate remaining // gas in the tank, and total gas for (int i = 0; i < n; i++) { currGas += gas[i] - cost[i]; totalGas += gas[i] - cost[i]; // If currGas is negative, circular tour can't // start with this index, so update it to next one if (currGas < 0) { currGas = 0; startIdx = i + 1; } } // No solution exists if (totalGas < 0) return -1; return startIdx; } static void Main() { int[] gas = { 1, 2, 3, 4, 5 }; int[] cost = { 3, 4, 5, 1, 2 }; Console.WriteLine(startStation(gas, cost)); } } // JavaScript program to find starting index of circular Tour // using greedy approach in one pass function startStation(gas, cost) { const n = gas.length; // Variables to track total and current remaining gas let totalGas = 0; let currGas = 0; let startIdx = 0; // Traverse through each station to calculate remaining // gas in the tank, and total gas for (let i = 0; i < n; i++) { currGas += gas[i] - cost[i]; totalGas += gas[i] - cost[i]; // If currGas is negative, circular tour can't // start with this index, so update it to next one if (currGas < 0) { currGas = 0; startIdx = i + 1; } } // No solution exists if (totalGas < 0) return -1; return startIdx; } // Driver Code const gas = [1, 2, 3, 4, 5]; const cost = [3, 4, 5, 1, 2]; console.log(startStation(gas, cost)); Similar Reads
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