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Commit bd9def9

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feat: add solutions to lc problems: No.2240,2241 (doocs#3335)
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5 files changed

+134
-3
lines changed

‎solution/2200-2299/2240.Number of Ways to Buy Pens and Pencils/README.md‎

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### 方法一:枚举
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我们可以枚举购买钢笔的数量 $x,ドル对于每个 $x,ドル我们最多可以购买铅笔的数量为 $\frac{total - x \times cost1}{cost2},ドル那么数量加 1ドル$ 即为 $x$ 的方案数。我们累加所有的 $x$ 的方案数,即为答案。
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我们可以枚举购买钢笔的数量 $x,ドル对于每个 $x,ドル我们最多可以购买铅笔的数量为 $\frac{\textit{total} - x \times \textit{cost1}}{\textit{cost2}},ドル那么数量加 1ドル$ 即为 $x$ 的方案数。我们累加所有的 $x$ 的方案数,即为答案。
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时间复杂度 $O(\frac{total}{cost1}),ドル空间复杂度 $O(1)$。
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‎solution/2200-2299/2240.Number of Ways to Buy Pens and Pencils/README_EN.md‎

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@@ -57,7 +57,11 @@ The total number of ways to buy pens and pencils is 5 + 3 + 1 = 9.
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<!-- solution:start -->
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### Solution 1
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### Solution 1: Enumeration
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We can enumerate the number of pens to buy, denoted as $x$. For each $x,ドル the maximum number of pencils we can buy is $\frac{\textit{total} - x \times \textit{cost1}}{\textit{cost2}}$. The number of ways for each $x$ is this value plus 1. We sum up the number of ways for all $x$ to get the answer.
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The time complexity is $O(\frac{\textit{total}}{\textit{cost1}}),ドル and the space complexity is $O(1)$.
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<!-- tabs:start -->
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‎solution/2200-2299/2241.Design an ATM Machine/README.md‎

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@@ -239,6 +239,48 @@ func (this *ATM) Withdraw(amount int) []int {
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*/
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```
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#### TypeScript
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```ts
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class ATM {
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private cnt: number[];
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private d: number[];
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constructor() {
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this.cnt = [0, 0, 0, 0, 0];
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this.d = [20, 50, 100, 200, 500];
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}
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deposit(banknotesCount: number[]): void {
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for (let i = 0; i < banknotesCount.length; i++) {
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this.cnt[i] += banknotesCount[i];
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}
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}
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withdraw(amount: number): number[] {
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let ans = [0, 0, 0, 0, 0];
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for (let i = 4; i >= 0; i--) {
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ans[i] = Math.min(Math.floor(amount / this.d[i]), this.cnt[i]);
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amount -= ans[i] * this.d[i];
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}
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if (amount > 0) {
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return [-1];
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}
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for (let i = 0; i < ans.length; i++) {
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this.cnt[i] -= ans[i];
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}
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return ans;
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}
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}
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/**
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* Your ATM object will be instantiated and called as such:
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* var obj = new ATM()
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* obj.deposit(banknotesCount)
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* var param_2 = obj.withdraw(amount)
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*/
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```
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<!-- tabs:end -->
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<!-- solution:end -->

‎solution/2200-2299/2241.Design an ATM Machine/README_EN.md‎

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@@ -82,7 +82,13 @@ atm.withdraw(550); // Returns [0,1,0,0,1]. The machine uses 1 50ドル banknot
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<!-- solution:start -->
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### Solution 1
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### Solution 1: Simulation
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We use an array $d$ to record the denominations of the bills and an array $cnt$ to record the number of bills for each denomination.
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For the `deposit` operation, we only need to add the number of bills for the corresponding denomination. The time complexity is $O(1)$.
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For the `withdraw` operation, we enumerate the bills from largest to smallest denomination, taking out as many bills as possible without exceeding the $amount$. Then, we subtract the total value of the withdrawn bills from $amount$. If $amount$ is still greater than 0ドル$ at the end, it means it's not possible to withdraw the $amount$ with the available bills, and we return $-1$. Otherwise, we return the number of bills withdrawn. The time complexity is $O(1)$.
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<!-- tabs:start -->
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*/
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```
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#### TypeScript
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```ts
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class ATM {
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private cnt: number[];
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private d: number[];
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constructor() {
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this.cnt = [0, 0, 0, 0, 0];
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this.d = [20, 50, 100, 200, 500];
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}
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deposit(banknotesCount: number[]): void {
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for (let i = 0; i < banknotesCount.length; i++) {
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this.cnt[i] += banknotesCount[i];
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}
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}
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withdraw(amount: number): number[] {
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let ans = [0, 0, 0, 0, 0];
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for (let i = 4; i >= 0; i--) {
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ans[i] = Math.min(Math.floor(amount / this.d[i]), this.cnt[i]);
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amount -= ans[i] * this.d[i];
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}
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if (amount > 0) {
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return [-1];
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}
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for (let i = 0; i < ans.length; i++) {
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this.cnt[i] -= ans[i];
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}
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return ans;
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}
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}
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/**
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* Your ATM object will be instantiated and called as such:
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* var obj = new ATM()
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* obj.deposit(banknotesCount)
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* var param_2 = obj.withdraw(amount)
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*/
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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class ATM {
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private cnt: number[];
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private d: number[];
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constructor() {
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this.cnt = [0, 0, 0, 0, 0];
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this.d = [20, 50, 100, 200, 500];
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}
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deposit(banknotesCount: number[]): void {
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for (let i = 0; i < banknotesCount.length; i++) {
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this.cnt[i] += banknotesCount[i];
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}
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}
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withdraw(amount: number): number[] {
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let ans = [0, 0, 0, 0, 0];
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for (let i = 4; i >= 0; i--) {
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ans[i] = Math.min(Math.floor(amount / this.d[i]), this.cnt[i]);
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amount -= ans[i] * this.d[i];
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}
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if (amount > 0) {
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return [-1];
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}
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for (let i = 0; i < ans.length; i++) {
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this.cnt[i] -= ans[i];
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}
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return ans;
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}
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}
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/**
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* Your ATM object will be instantiated and called as such:
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* var obj = new ATM()
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* obj.deposit(banknotesCount)
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* var param_2 = obj.withdraw(amount)
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*/

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