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860. 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 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.

Solutions

Solution 1

Thinking

Bills are only \(5\), \(10\), and \(20\), and change must come from cash already taken. \(n\le 10^5\), so greedy: a \(20\) prefers one \(10\) plus one \(5\), otherwise three \(5\)s.

Track counts of \(5\)s and \(10\)s; a negative \(5\) count fails. The full history is unnecessary.

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class Solution:
    def lemonadeChange(self, bills: List[int]) -> bool:
        five = ten = 0
        for v in bills:
            if v == 5:
                five += 1
            elif v == 10:
                ten += 1
                five -= 1
            else:
                if ten:
                    ten -= 1
                    five -= 1
                else:
                    five -= 3
            if five < 0:
                return False
        return True

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

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

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func lemonadeChange(bills []int) bool {
    five, ten := 0, 0
    for _, v := range bills {
        if v == 5 {
            five++
        } else if v == 10 {
            ten++
            five--
        } else {
            if ten > 0 {
                ten--
                five--
            } else {
                five -= 3
            }
        }
        if five < 0 {
            return false
        }
    }
    return true
}

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function lemonadeChange(bills: number[]): boolean {
    let [five, ten] = [0, 0];
    for (const x of bills) {
        switch (x) {
            case 5:
                five++;
                break;
            case 10:
                five--;
                ten++;
                break;
            case 20:
                if (ten) {
                    ten--;
                    five--;
                } else {
                    five -= 3;
                }
                break;
        }

        if (five < 0) {
            return false;
        }
    }
    return true;
}

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function lemonadeChange(bills) {
    let [five, ten] = [0, 0];
    for (const x of bills) {
        switch (x) {
            case 5:
                five++;
                break;
            case 10:
                five--;
                ten++;
                break;
            case 20:
                if (ten) {
                    ten--;
                    five--;
                } else {
                    five -= 3;
                }
                break;
        }

        if (five < 0) {
            return false;
        }
    }
    return true;
}

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impl Solution {
    pub fn lemonade_change(bills: Vec<i32>) -> bool {
        let (mut five, mut ten) = (0, 0);
        for bill in bills.iter() {
            match bill {
                5 => {
                    five += 1;
                }
                10 => {
                    five -= 1;
                    ten += 1;
                }
                _ => {
                    if ten != 0 {
                        ten -= 1;
                        five -= 1;
                    } else {
                        five -= 3;
                    }
                }
            }

            if five < 0 {
                return false;
            }
        }
        true
    }
}

Solution 2: One-liner

Thinking

Method 1 already covers every change rule. Method 2 folds the same two counters into one \(\textit{every}\) scan, branching on denomination with xor tests.

The meaning is unchanged; only the implementation is a one-liner, still failing when the \(5\) count goes negative.

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const lemonadeChange = (bills: number[], f = 0, t = 0): boolean =>
    bills.every(
        x => (
            (!(x ^ 5) && ++f) ||
                (!(x ^ 10) && (--f, ++t)) ||
                (!(x ^ 20) && (t ? (f--, t--) : (f -= 3), 1)),
            f >= 0
        ),
    );

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const lemonadeChange = (bills, f = 0, t = 0) =>
    bills.every(
        x => (
            (!(x ^ 5) && ++f) ||
                (!(x ^ 10) && (--f, ++t)) ||
                (!(x ^ 20) && (t ? (f--, t--) : (f -= 3), 1)),
            f >= 0
        ),
    );

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