Portfolio

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 <= 10^5

0 <= gas[i], cost[i] <= 10^4

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효성

n^2인거 예상은 했지만 고집부려서 시간 초과당한 거 결국 봤네요 ^^..


var canCompleteCircuit = function(gas, cost) {
    const n = gas.length;
    let maxDiffIdxList = [];

    for(let i = 0; i < n; i++) {
        const diff = gas[i] - cost[i];
        if(diff >= 0) {
            maxDiffIdxList.push(i);
        }
    }
    for(let i = 0; i < maxDiffIdxList.length; i++) {
        const curIdx = maxDiffIdxList[i];
        const newGas = gas.slice(curIdx, n).concat(gas.slice(0, curIdx));
        const newCost = cost.slice(curIdx, n).concat(cost.slice(0, curIdx));

        if(canGo(newGas, newCost)) {
            return curIdx;
        }
    }

    return -1;
};

function canGo(gas, cost) {
    const n = gas.length;
    let curTank = gas[0];
    for(let i = 0; i < n; i++) {
        curTank -= cost[i];
        if(curTank < 0) {
            return false;
        }
        curTank += gas[i + 1];
    }
    return curTank < 0 ? false : true;
}

참고한 풀이 (엇 저희 이 문제 풀었네요..ㅎㅎ.. 복습 차원으로 낸 걸루..)


 var canCompleteCircuit = function(gas, cost) {
    let curTank = 0
    let totalTank = 0
    let pos = 0;
    for (let i = 0; i < gas.length; i++) {
        curTank += gas[i] - cost[i];
        totalTank += gas[i] - cost[i];

        if (curTank < 0) {
            curTank = 0;
            pos = i + 1;
        }
    }   

    return totalTank < 0 ? -1 : pos;
}

~~재영: 네~ 종아리 대시면 됩니다~! ^0^~~

재영

Previous: 110ms (Greedy, 굳이 세부적으로 따지자면 O(3 * N))

After: 78ms (used PrefixSum, O(N))


const canCompleteCircuit = (gas, cost) => {
  const prefixSumStack = [];

  let maxValue = -Infinity;
  let maxValueIndex = 0;

  let total = 0;
  
  for (let i = gas.length - 1; i >= 0; i -= 1) {
    const diff = gas[i] - cost[i];

    total += diff;

    if (total >= maxValue) {
      maxValue = total;
      maxValueIndex = i;
    }

    prefixSumStack.push(total);
  }

  return prefixSumStack.pop() < 0 ? -1 : maxValueIndex;
};

Gas Station

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