Lemonade Change

Description:

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 given 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 $i^{th}$ 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 bill. From the fifth customer, we give a $10 bill and a $5 bill. Since all customers received correct change, we return 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 change of $15 back because we only have two $10 bills. Since not every customer received correct change, the answer is false.

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Approaches:

1. Greedy Algorithm

When a customer pays us, we maintain a count of the $5 and $10 bills we have in hand (we don't need to count $20 bills because we can never use them to give change).

There are three scenarios based on the bill received:

1. Received $5: No change needed. Increment our $5 count.
2. Received $10: Requires $5 change. If we have a $5 bill, use it; otherwise, return false.
3. Received $20: Requires $15 change. This is where the greedy choice happens. We have two ways to make $15:

   • Option A (Optimal): Give one $10 bill and one $5 bill.

   • Option B: Give three $5 bills.

   Greedy Strategy: Always prefer Option A. A $5 bill is much more valuable/flexible than a $10 bill because a $5 bill can help give change for both $10 and $20 transactions, whereas a $10 bill can only help with $20 transactions.

• Time Complexity: $O(n)$ because we iterate through the bills array once.
• Space Complexity: $O(1)$ because we only use a few variables for the calculation, taking constant extra space.

Greedy Solutions:

C++

class Solution {
public:
    bool lemonadeChange(vector<int>& bills) {
        int five = 0, ten = 0;
        
        for (int bill : bills) {
            if (bill == 5) {
                five++;
            } else if (bill == 10) {
                if (five == 0) return false;
                five--;
                ten++;
            } else { // bill == 20
                if (ten > 0 && five > 0) {
                    ten--;
                    five--;
                } else if (five >= 3) {
                    five -= 3;
                } else {
                    return false;
                }
            }
        }
        return true;
    }
};

Java

class Solution {
    public boolean lemonadeChange(int[] bills) {
        int five = 0, ten = 0;
        
        for (int bill : bills) {
            if (bill == 5) {
                five++;
            } else if (bill == 10) {
                if (five == 0) return false;
                five--;
                ten++;
            } else { // bill == 20
                if (ten > 0 && five > 0) {
                    ten--;
                    five--;
                } else if (five >= 3) {
                    five -= 3;
                } else {
                    return false;
                }
            }
        }
        return true;
    }
}

Python

class Solution:
    def lemonadeChange(self, bills: List[int]) -> bool:
        five = ten = 0
        
        for bill in bills:
            if bill == 5:
                five += 1
            elif bill == 10:
                if not five:
                    return False
                five -= 1
                ten += 1
            else: # bill == 20
                if ten and five:
                    ten -= 1
                    five -= 1
                elif five >= 3:
                    five -= 3
                else:
                    return False
        return True

JavaScript

/**
 * @param {number[]} bills
 * @return {boolean}
 */
var lemonadeChange = function(bills) {
    let five = 0, ten = 0;
    
    for (let bill of bills) {
        if (bill === 5) {
            five++;
        } else if (bill === 10) {
            if (five === 0) return false;
            five--;
            ten++;
        } else { // bill === 20
            if (ten > 0 && five > 0) {
                ten--;
                five--;
            } else if (five >= 3) {
                five -= 3;
            } else {
                return false;
            }
        }
    }
    return true;
};