LeetCode - Lemonade Change (Javascript)

문제 설명

줄 서있는 손님들에게 돈을 받은 뒤, 거스름돈을 돌려줄 수 있는지 없는지 판별하는 문제다. (5달러, 10달러, 20달러만 있다고 가정)

At a lemonade stand, each lemonade costs $5. Customers are standing in a queue to buy from you and order one at a time (in the order specified by bills). Each customer will only buy one lemonade and pay with either a $5, $10, or $20 bill. You must provide the correct change to each customer so that the net transaction is that the customer pays $5.

Note that you do not have any change in hand at first.

Given an integer array bills where bills[i] is the bill the ith customer pays, return true if you can provide every customer with the correct change, or false otherwise.

Example 1:

Input: bills = [5,5,5,10,20]
Output: true
Explanation:
From the first 3 customers, we collect three $5 bills in order.
From the fourth customer, we collect a $10 bill and give back a $5.
From the fifth customer, we give a $10 bill and a $5 bill.
Since all customers got correct change, we output true.

Example 2:

Input: bills = [5,5,10,10,20]
Output: false
Explanation:
From the first two customers in order, we collect two $5 bills.
For the next two customers in order, we collect a $10 bill and give back a $5 bill.
For the last customer, we can not give the change of $15 back because we only have two $10 bills.
Since not every customer received the correct change, the answer is false.

Constraints:

• 1 <= bills.length <= 105
• bills[i] is either 5, 10, or 20.

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소스코드

한 번에 성공했다.
시간 효율성도 좋게 나왔는데, 그래도 조금 마음에 걸리는 것은 너무 하드코딩을 했다는 것이다 ㅜ

const lemonadeChange = (bills) => {
  const changeObj = {};
  if (bills[0] > 5) return false;
  for (let i = 0; i < bills.length; i++) {
    changeObj[bills[i]] = changeObj[bills[i]] + 1 || 1;
    if (bills[i] === 5) continue;
    if (bills[i] === 10) {
      if (!changeObj[5]) return false;
      changeObj[5]--;
    }
    if (bills[i] === 20) {
      if (!changeObj[5] || (!changeObj[10] && changeObj[5] < 3)) return false;
      if (changeObj[10]) {
        changeObj[10]--;
        changeObj[5]--;
      } else {
        changeObj[5] -= 3;
      }
    }
  }
  return true;
};

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