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2241. Design an ATM Machine
Description
There is an ATM machine that stores banknotes of 5
denominations: 20
, 50
, 100
, 200
, and 500
dollars. Initially the ATM is empty. The user can use the machine to deposit or withdraw any amount of money.
When withdrawing, the machine prioritizes using banknotes of larger values.
- For example, if you want to withdraw
300
and there are2
50
banknotes,1
100
banknote, and1
200
banknote, then the machine will use the100
and200
banknotes. - However, if you try to withdraw
600
and there are3
200
banknotes and1
500
banknote, then the withdraw request will be rejected because the machine will first try to use the500
banknote and then be unable to use banknotes to complete the remaining100
. Note that the machine is not allowed to use the200
banknotes instead of the500
banknote.
Implement the ATM class:
ATM()
Initializes the ATM object.void deposit(int[] banknotesCount)
Deposits new banknotes in the order20
,50
,100
,200
, and500
.int[] withdraw(int amount)
Returns an array of length5
of the number of banknotes that will be handed to the user in the order20
,50
,100
,200
, and500
, and update the number of banknotes in the ATM after withdrawing. Returns[-1]
if it is not possible (do not withdraw any banknotes in this case).
Example 1:
Input ["ATM", "deposit", "withdraw", "deposit", "withdraw", "withdraw"] [[], [[0,0,1,2,1]], [600], [[0,1,0,1,1]], [600], [550]] Output [null, null, [0,0,1,0,1], null, [-1], [0,1,0,0,1]] Explanation ATM atm = new ATM(); atm.deposit([0,0,1,2,1]); // Deposits 1 100 banknote, 2 200 banknotes, // and 1 500 banknote. atm.withdraw(600); // Returns [0,0,1,0,1]. The machine uses 1 100 banknote // and 1 500 banknote. The banknotes left over in the // machine are [0,0,0,2,0]. atm.deposit([0,1,0,1,1]); // Deposits 1 50, 200, and 500 banknote. // The banknotes in the machine are now [0,1,0,3,1]. atm.withdraw(600); // Returns [-1]. The machine will try to use a 500 banknote // and then be unable to complete the remaining 100, // so the withdraw request will be rejected. // Since the request is rejected, the number of banknotes // in the machine is not modified. atm.withdraw(550); // Returns [0,1,0,0,1]. The machine uses 1 50 banknote // and 1 500 banknote.
Constraints:
banknotesCount.length == 5
0 <= banknotesCount[i] <= 109
1 <= amount <= 109
- At most
5000
calls in total will be made towithdraw
anddeposit
. - At least one call will be made to each function
withdraw
anddeposit
.
Solutions
-
class ATM { private long[] cnt = new long[5]; private int[] d = {20, 50, 100, 200, 500}; public ATM() { } public void deposit(int[] banknotesCount) { for (int i = 0; i < banknotesCount.length; ++i) { cnt[i] += banknotesCount[i]; } } public int[] withdraw(int amount) { int[] ans = new int[5]; for (int i = 4; i >= 0; --i) { ans[i] = (int) Math.min(amount / d[i], cnt[i]); amount -= ans[i] * d[i]; } if (amount > 0) { return new int[] {-1}; } for (int i = 0; i < 5; ++i) { cnt[i] -= ans[i]; } return ans; } } /** * Your ATM object will be instantiated and called as such: * ATM obj = new ATM(); * obj.deposit(banknotesCount); * int[] param_2 = obj.withdraw(amount); */
-
class ATM { public: ATM() { } void deposit(vector<int> banknotesCount) { for (int i = 0; i < banknotesCount.size(); ++i) { cnt[i] += banknotesCount[i]; } } vector<int> withdraw(int amount) { vector<int> ans(5); for (int i = 4; ~i; --i) { ans[i] = min(1ll * amount / d[i], cnt[i]); amount -= ans[i] * d[i]; } if (amount > 0) { return {-1}; } for (int i = 0; i < 5; ++i) { cnt[i] -= ans[i]; } return ans; } private: long long cnt[5] = {0}; int d[5] = {20, 50, 100, 200, 500}; }; /** * Your ATM object will be instantiated and called as such: * ATM* obj = new ATM(); * obj->deposit(banknotesCount); * vector<int> param_2 = obj->withdraw(amount); */
-
class ATM: def __init__(self): self.cnt = [0] * 5 self.d = [20, 50, 100, 200, 500] def deposit(self, banknotesCount: List[int]) -> None: for i, v in enumerate(banknotesCount): self.cnt[i] += v def withdraw(self, amount: int) -> List[int]: ans = [0] * 5 for i in range(4, -1, -1): ans[i] = min(amount // self.d[i], self.cnt[i]) amount -= ans[i] * self.d[i] if amount > 0: return [-1] for i, v in enumerate(ans): self.cnt[i] -= v return ans # Your ATM object will be instantiated and called as such: # obj = ATM() # obj.deposit(banknotesCount) # param_2 = obj.withdraw(amount)
-
type ATM struct { d [5]int cnt [5]int } func Constructor() ATM { return ATM{[5]int{20, 50, 100, 200, 500}, [5]int{} } } func (this *ATM) Deposit(banknotesCount []int) { for i, v := range banknotesCount { this.cnt[i] += v } } func (this *ATM) Withdraw(amount int) []int { ans := make([]int, 5) for i := 4; i >= 0; i-- { ans[i] = min(amount/this.d[i], this.cnt[i]) amount -= ans[i] * this.d[i] } if amount > 0 { return []int{-1} } for i, v := range ans { this.cnt[i] -= v } return ans } /** * Your ATM object will be instantiated and called as such: * obj := Constructor(); * obj.Deposit(banknotesCount); * param_2 := obj.Withdraw(amount); */
-
class ATM { private cnt: number[]; private d: number[]; constructor() { this.cnt = [0, 0, 0, 0, 0]; this.d = [20, 50, 100, 200, 500]; } deposit(banknotesCount: number[]): void { for (let i = 0; i < banknotesCount.length; i++) { this.cnt[i] += banknotesCount[i]; } } withdraw(amount: number): number[] { let ans = [0, 0, 0, 0, 0]; for (let i = 4; i >= 0; i--) { ans[i] = Math.min(Math.floor(amount / this.d[i]), this.cnt[i]); amount -= ans[i] * this.d[i]; } if (amount > 0) { return [-1]; } for (let i = 0; i < ans.length; i++) { this.cnt[i] -= ans[i]; } return ans; } } /** * Your ATM object will be instantiated and called as such: * var obj = new ATM() * obj.deposit(banknotesCount) * var param_2 = obj.withdraw(amount) */