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Home Code Solutions

Gas Station – LeetCode Solution

Gas Station - LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

BhautikBhalala by BhautikBhalala
September 5, 2022
Reading Time: 2 mins read
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Leetcode All Problems Solutions

Leetcode All Problems Solutions

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Table of Contents

  • Gas Station – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
  • C++ Gas Station LeetCode Solution
  • Java Gas Station LeetCode Solution
  • Python 3 Gas Station LeetCode Solution
    • Leave a comment below
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Gas Station – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

There are n gas stations along a circular route, where the amount of gas at the ith station is gas[i].

You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from the ith station to its next (i + 1)th station. You begin the journey with an empty tank at one of the gas stations.

Given two integer arrays gas and cost, return the starting gas station’s index if you can travel around the circuit once in the clockwise direction, otherwise return -1. If there exists a solution, it is guaranteed to be unique

 

Example 1:

Input: gas = [1,2,3,4,5], cost = [3,4,5,1,2]
Output: 3
Explanation:
Start at station 3 (index 3) and fill up with 4 unit of gas. Your tank = 0 + 4 = 4
Travel to station 4. Your tank = 4 - 1 + 5 = 8
Travel to station 0. Your tank = 8 - 2 + 1 = 7
Travel to station 1. Your tank = 7 - 3 + 2 = 6
Travel to station 2. Your tank = 6 - 4 + 3 = 5
Travel to station 3. The cost is 5. Your gas is just enough to travel back to station 3.
Therefore, return 3 as the starting index.

Example 2:

Input: gas = [2,3,4], cost = [3,4,3]
Output: -1
Explanation:
You can't start at station 0 or 1, as there is not enough gas to travel to the next station.
Let's start at station 2 and fill up with 4 unit of gas. Your tank = 0 + 4 = 4
Travel to station 0. Your tank = 4 - 3 + 2 = 3
Travel to station 1. Your tank = 3 - 3 + 3 = 3
You cannot travel back to station 2, as it requires 4 unit of gas but you only have 3.
Therefore, you can't travel around the circuit once no matter where you start.

 

Constraints:

  • n == gas.length == cost.length
  • 1 <= n <= 105
  • 0 <= gas[i], cost[i] <= 104

C++ Gas Station LeetCode Solution


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class Solution {
public:
	int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
		int gasReq = accumulate(gas.begin(), gas.end(), 0);
		int costReq = accumulate(cost.begin(), cost.end(), 0);
		if (costReq > gasReq) return -1;
		int ans = 0;
		int currS = 0;
		for (int i = 0; i < gas.size(); ++i) {
			currS += gas[i] - cost[i];
			if (currS < 0) {
				currS = 0;
				ans = i + 1;
			}
		}
		return ans;
	}
};

Java Gas Station LeetCode Solution


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class Solution {
    public int canCompleteCircuit(int[] gas, int[] costs) {
        int totalCost = 0;
        int totalFuel = 0;
        int current = 0;
        int start = 0;
        
        for(int fuel : gas){
            totalFuel += fuel;
        }
        for(int cost : costs){
            totalCost += cost;
        }
        
        if(totalFuel < totalCost){
         return -1;   
        }
        
        
        for(int index=0;index<gas.length;index++){
            currentfuel += (gas[index] - costs[index]);
            if(currentfuel < 0){
                start = index+1;
                currentfuel = 0;
            }
        }
        
        return start;
    }
}

 



Python 3 Gas Station LeetCode Solution


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 def canCompleteCircuit(self, gas: List[int], cost: List[int]) -> int:
        if sum(cost) > sum(gas): return -1
        diff = list(accumulate([x - y for x, y in zip(gas, cost)]))
        return (diff.index(min(diff)) + 1) % len(diff)



 

Array-1180
String-562
Hash Table-412
Dynamic Programming-390
Math-368
Sorting-264
Greedy-257
Depth-First Search-256
Database-215
Breadth-First Search-200
Tree-195
Binary Search-191
Matrix-176
Binary Tree-160
Two Pointers-151
Bit Manipulation-140
Stack-133
Heap (Priority Queue)-117
Design-116
Graph-108
Simulation-103
Prefix Sum-96
Backtracking-92
Counting-86
Sliding Window-73
Linked List-69
Union Find-66
Ordered Set-48
Monotonic Stack-47
Recursion-43
Trie-41
Binary Search Tree-40
Divide and Conquer-40
Enumeration-39
Bitmask-37
Queue-33
Memoization-32
Topological Sort-31
Geometry-30
Segment Tree-27
Game Theory-24
Hash Function-24
Binary Indexed Tree-21
Interactive-18
Data Stream-17
String Matching-17
Rolling Hash-17
Shortest Path-16
Number Theory-16
Combinatorics-15
Randomized-12
Monotonic Queue-9
Iterator-9
Merge Sort-9
Concurrency-9
Doubly-Linked List-8
Brainteaser-8
Probability and Statistics-7
Quickselect-7
Bucket Sort-6
Suffix Array-6
Minimum Spanning Tree-5
Counting Sort-5
Shell-4
Line Sweep-4
Reservoir Sampling-4
Eulerian Circuit-3
Radix Sort-3
Strongly Connected Componen-t2
Rejection Sampling-2
Biconnected Component-1

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Tags: Cc++14full solutionGas StationGojavajava 15java 7java 8java8javascriptkotlinLeetCodeLeetCodeSolutionpypy 3Python 2python 3rubyrustSolution
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