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

Author: yanglbme

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

@@ -59,7 +59,7 @@ tags:
 
 ### 方法一：枚举
 
-我们可以枚举购买钢笔的数量 $x$，对于每个 $x$，我们最多可以购买铅笔的数量为 $\frac{total - x \times cost1}{cost2}$，那么数量加 $1$ 即为 $x$ 的方案数。我们累加所有的 $x$ 的方案数，即为答案。
+我们可以枚举购买钢笔的数量 $x$，对于每个 $x$，我们最多可以购买铅笔的数量为 $\frac{\textit{total} - x \times \textit{cost1}}{\textit{cost2}}$，那么数量加 $1$ 即为 $x$ 的方案数。我们累加所有的 $x$ 的方案数，即为答案。
 
 时间复杂度 $O(\frac{total}{cost1})$，空间复杂度 $O(1)$。
 

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

@@ -57,7 +57,11 @@ The total number of ways to buy pens and pencils is 5 + 3 + 1 = 9.
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Enumeration
+
+We can enumerate the number of pens to buy, denoted as $x$. For each $x$, 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.
+
+The time complexity is $O(\frac{\textit{total}}{\textit{cost1}})$, and the space complexity is $O(1)$.
 
 <!-- tabs:start -->
 

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

@@ -239,6 +239,48 @@ func (this *ATM) Withdraw(amount int) []int {
  */
 ```
 
+#### TypeScript
+
+```ts
+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)
+ */
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

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

@@ -82,7 +82,13 @@ atm.withdraw(550);        // Returns [0,1,0,0,1]. The machine uses 1 $50 banknot
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Simulation
+
+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.
+
+For the `deposit` operation, we only need to add the number of bills for the corresponding denomination. The time complexity is $O(1)$.
+
+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 $0$ 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 $O(1)$.
 
 <!-- tabs:start -->
 
@@ -239,6 +245,48 @@ func (this *ATM) Withdraw(amount int) []int {
  */
 ```
 
+#### TypeScript
+
+```ts
+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)
+ */
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/2200-2299/2241.Design an ATM Machine/Solution.ts

@@ -0,0 +1,37 @@
+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)
+ */