How to Solve the Lemonade Change Problem by Simulation Algorithm?

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 don’t have any change in hand at first.

Return true if and only if you can provide every customer with correct changes.

Example 1:
Input: [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: [5,5,10]
Output: true

Example 3:
Input: [10,10]
Output: false

Example 4:
Input: [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’t give change of $15 back because we only have two $10 bills.
Since not every customer received correct change, the answer is false.

Note:
0 <= bills.length <= 10000
bills[i] will be either 5, 10, or 20

The problem can be simulated by the process as each customer can only have three situations – 5, 10 or 20 note.

We have two counters that storing the current number of notes for $5 and $10. If this customer has $10, we need to check if we have $5 note, update the counter and continue, otherwise fail with false.

Similarly, with $20, we need to give change $10 + $5 or, $5 x 3, in the preferred order (Greedy Algorithm).

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24

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

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

O(N) complexity where N is the number of the customers and O(1) constant space complexity as we only need two variables.

–EOF (The Ultimate Computing & Technology Blog) —