LeetCode 0860. Lemonade Change Solution in Java, C++, Python & More | Explanation + Code

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 a5.
From the fifth customer, we give a 10 bill and a5 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 a5 bill.
For the last customer, we can not give the change of 15 back because we only have two10 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

PythonJavaC++GoTypeScriptJavaScriptRust

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(code-box)

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; } }(code-box)

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; } };(code-box)

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 }(code-box)

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; }(code-box)

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; }(code-box)

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 } }(code-box)

Solution 2: One-liner

TypeScriptJavaScript

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 ), );(code-box)

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 ), );(code-box)