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134.gas-station.js
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134.gas-station.js
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/*
* @lc app=leetcode id=134 lang=javascript
*
* [134] Gas Station
*
* https://leetcode.com/problems/gas-station/description/
*
* algorithms
* Medium (35.80%)
* Likes: 1060
* Dislikes: 331
* Total Accepted: 170.9K
* Total Submissions: 477.4K
* Testcase Example: '[1,2,3,4,5]\n[3,4,5,1,2]'
*
* There are N gas stations along a circular route, where the amount of gas at
* station i is gas[i].
*
* You have a car with an unlimited gas tank and it costs cost[i] of gas to
* travel from station i to its next station (i+1). You begin the journey with
* an empty tank at one of the gas stations.
*
* Return the starting gas station's index if you can travel around the circuit
* once in the clockwise direction, otherwise return -1.
*
* Note:
*
*
* If there exists a solution, it is guaranteed to be unique.
* Both input arrays are non-empty and have the same length.
* Each element in the input arrays is a non-negative integer.
*
*
* 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.
*
*
*/
// @lc code=start
/**
* @param {number[]} gas
* @param {number[]} cost
* @return {number}
*/
const canCompleteCircuit = function(gas, cost) {
let sum = 0, start = 0, c = 0;
for (let i = 0; i < gas.length; i++) {
sum = sum + gas[i] - cost[i];
if(sum < 0) {
start = i + 1;
c += sum;
sum = 0;
}
}
return sum + c >=0 ? start : -1;
};
// @lc code=end