860. Lemonade Change

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.

Approach 01:

class Solution {
public:
    bool lemonadeChange(vector<int>& bills) {
        int fiveDollarCount = 0;
        int tenDollarCount = 0;

        for (const int bill : bills) {
            if (bill == 5) {
                ++fiveDollarCount;
            } else if (bill == 10) {
                --fiveDollarCount;
                ++tenDollarCount;
            } else { // bill == 20
                if (tenDollarCount > 0) {
                    --tenDollarCount;
                    --fiveDollarCount;
                } else {
                    fiveDollarCount -= 3;
                }
            }
            if (fiveDollarCount < 0)
                return false;
        }

        return true;
    }
};

class Solution:
    def lemonadeChange(self, bills: List[int]) -> bool:
        fiveDollarCount = 0
        tenDollarCount = 0

        for bill in bills:
            if bill == 5:
                fiveDollarCount += 1
            elif bill == 10:
                fiveDollarCount -= 1
                tenDollarCount += 1
            else:  # bill == 20
                if tenDollarCount > 0:
                    tenDollarCount -= 1
                    fiveDollarCount -= 1
                else:
                    fiveDollarCount -= 3

            if fiveDollarCount < 0:
                return False

        return True

Time Complexity

• Iterating Through Bills:

  The function iterates through each bill in the bills vector exactly once. Each iteration involves a constant amount of work (checking conditions and updating counters), making the time complexity \( O(n) \), where \( n \) is the number of bills.

Space Complexity

• Space Usage:

  The function uses a fixed amount of additional space for storing two integer counters (fiveDollarCount and tenDollarCount). This results in a space complexity of \( O(1) \), indicating constant space usage.