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134. Gas Station
Description
There are n
gas stations along a circular route, where the amount of gas at the i^{th}
station is gas[i]
.
You have a car with an unlimited gas tank and it costs cost[i]
of gas to travel from the i^{th}
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 <= 10^{5}
0 <= gas[i], cost[i] <= 10^{4}
Solutions

class Solution { public int canCompleteCircuit(int[] gas, int[] cost) { int n = gas.length; int i = n  1, j = n  1; int cnt = 0, s = 0; while (cnt < n) { s += gas[j]  cost[j]; ++cnt; j = (j + 1) % n; while (s < 0 && cnt < n) { i; s += gas[i]  cost[i]; ++cnt; } } return s < 0 ? 1 : i; } }

class Solution { public: int canCompleteCircuit(vector<int>& gas, vector<int>& cost) { int n = gas.size(); int i = n  1, j = n  1; int cnt = 0, s = 0; while (cnt < n) { s += gas[j]  cost[j]; ++cnt; j = (j + 1) % n; while (s < 0 && cnt < n) { i; s += gas[i]  cost[i]; ++cnt; } } return s < 0 ? 1 : i; } };

class Solution: def canCompleteCircuit(self, gas: List[int], cost: List[int]) > int: n = len(gas) i = j = n  1 cnt = s = 0 while cnt < n: s += gas[j]  cost[j] cnt += 1 j = (j + 1) % n while s < 0 and cnt < n: i = 1 s += gas[i]  cost[i] cnt += 1 return 1 if s < 0 else i

func canCompleteCircuit(gas []int, cost []int) int { n := len(gas) i, j := n1, n1 cnt, s := 0, 0 for cnt < n { s += gas[j]  cost[j] cnt++ j = (j + 1) % n for s < 0 && cnt < n { i s += gas[i]  cost[i] cnt++ } } if s < 0 { return 1 } return i }

function canCompleteCircuit(gas: number[], cost: number[]): number { const n = gas.length; let i = n  1; let j = n  1; let s = 0; let cnt = 0; while (cnt < n) { s += gas[j]  cost[j]; ++cnt; j = (j + 1) % n; while (s < 0 && cnt < n) { i; s += gas[i]  cost[i]; ++cnt; } } return s < 0 ? 1 : i; }

public class Solution { public int CanCompleteCircuit(int[] gas, int[] cost) { int n = gas.Length; int i = n  1, j = n  1; int s = 0, cnt = 0; while (cnt < n) { s += gas[j]  cost[j]; ++cnt; j = (j + 1) % n; while (s < 0 && cnt < n) { i; s += gas[i]  cost[i]; ++cnt; } } return s < 0 ? 1 : i; } }