feat: add new lc problems

solution/2800-2899/2806.Account Balance After Rounded Purchase/README.md

@@ -0,0 +1,157 @@
+# [2806. 取整购买后的账户余额](https://leetcode.cn/problems/account-balance-after-rounded-purchase)
+
+[English Version](/solution/2800-2899/2806.Account%20Balance%20After%20Rounded%20Purchase/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>一开始，你的银行账户里有&nbsp;<code>100</code>&nbsp;块钱。</p>
+
+<p>给你一个整数<code>purchaseAmount</code>&nbsp;，它表示你在一次购买中愿意支出的金额。</p>
+
+<p>在一个商店里，你进行一次购买，实际支出的金额会向 <strong>最近</strong>&nbsp;的&nbsp;<code>10</code>&nbsp;的 <strong>倍数</strong>&nbsp;取整。换句话说，你实际会支付一个&nbsp;<strong>非负</strong>&nbsp;金额&nbsp;<code>roundedAmount</code>&nbsp;，满足&nbsp;<code>roundedAmount</code>&nbsp;是&nbsp;<code>10</code>&nbsp;的倍数且&nbsp;<code>abs(roundedAmount - purchaseAmount)</code>&nbsp;的值 <strong>最小</strong>&nbsp;。</p>
+
+<p>如果存在多于一个最接近的 <code>10</code>&nbsp;的倍数，<strong>较大的倍数</strong>&nbsp;是你的实际支出金额。</p>
+
+<p>请你返回一个整数，表示你在愿意支出金额为<em>&nbsp;</em><code>purchaseAmount</code><em>&nbsp;</em>块钱的前提下，购买之后剩下的余额。</p>
+
+<p><strong>注意：</strong> <code>0</code>&nbsp;也是&nbsp;<code>10</code>&nbsp;的倍数。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre><b>输入：</b>purchaseAmount = 9
+<b>输出：</b>90
+<b>解释：</b>这个例子中，最接近 9 的 10 的倍数是 10 。所以你的账户余额为 100 - 10 = 90 。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre><b>输入：</b>purchaseAmount = 15
+<b>输出：</b>80
+<b>解释：</b>这个例子中，有 2 个最接近 15 的 10 的倍数：10 和 20，较大的数 20 是你的实际开销。
+所以你的账户余额为 100 - 20 = 80 。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>0 &lt;= purchaseAmount &lt;= 100</code></li>
+</ul>
+
+## 解法
+
+<!-- 这里可写通用的实现逻辑 -->
+
+**方法一：枚举 + 模拟**
+
+我们在 $[0, 100]$ 的范围内枚举所有的 $10$ 的倍数，然后找到与 `purchaseAmount` 最接近的那个数，记为 $x$，那么答案就是 $100 - x$。
+
+时间复杂度 $O(1)$，空间复杂度 $O(1)$。
+
+<!-- tabs:start -->
+
+### **Python3**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```python
+class Solution:
+    def accountBalanceAfterPurchase(self, purchaseAmount: int) -> int:
+        diff, x = 100, 0
+        for y in range(100, -1, -10):
+            if (t := abs(y - purchaseAmount)) < diff:
+                diff = t
+                x = y
+        return 100 - x
+```
+
+### **Java**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+class Solution {
+    public int accountBalanceAfterPurchase(int purchaseAmount) {
+        int diff = 100, x = 0;
+        for (int y = 100; y >= 0; y -= 10) {
+            int t = Math.abs(y - purchaseAmount);
+            if (t < diff) {
+                diff = t;
+                x = y;
+            }
+        }
+        return 100 - x;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int accountBalanceAfterPurchase(int purchaseAmount) {
+        int diff = 100, x = 0;
+        for (int y = 100; y >= 0; y -= 10) {
+            int t = abs(y - purchaseAmount);
+            if (t < diff) {
+                diff = t;
+                x = y;
+            }
+        }
+        return 100 - x;
+    }
+};
+```
+
+### **Go**
+
+```go
+func accountBalanceAfterPurchase(purchaseAmount int) int {
+	diff, x := 100, 0
+	for y := 100; y >= 0; y -= 10 {
+		t := abs(y - purchaseAmount)
+		if t < diff {
+			diff = t
+			x = y
+		}
+	}
+	return 100 - x
+}
+
+func abs(x int) int {
+	if x < 0 {
+		return -x
+	}
+	return x
+}
+```
+
+### **TypeScript**
+
+```ts
+function accountBalanceAfterPurchase(purchaseAmount: number): number {
+    let [diff, x] = [100, 0];
+    for (let y = 100; y >= 0; y -= 10) {
+        const t = Math.abs(y - purchaseAmount);
+        if (t < diff) {
+            diff = t;
+            x = y;
+        }
+    }
+    return 100 - x;
+}
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2800-2899/2806.Account Balance After Rounded Purchase/README_EN.md

@@ -0,0 +1,143 @@
+# [2806. Account Balance After Rounded Purchase](https://leetcode.com/problems/account-balance-after-rounded-purchase)
+
+[中文文档](/solution/2800-2899/2806.Account%20Balance%20After%20Rounded%20Purchase/README.md)
+
+## Description
+
+<p>Initially, you have a bank account balance of <code>100</code> dollars.</p>
+
+<p>You are given an integer <code>purchaseAmount</code> representing the amount you will spend on a purchase in dollars.</p>
+
+<p>At the store where you will make the purchase, the purchase amount is rounded to the <strong>nearest multiple</strong> of <code>10</code>. In other words, you pay a <strong>non-negative</strong> amount, <code>roundedAmount</code>, such that <code>roundedAmount</code> is a multiple of <code>10</code> and <code>abs(roundedAmount - purchaseAmount)</code> is <strong>minimized</strong>.</p>
+
+<p>If there is more than one nearest multiple of <code>10</code>, the <strong>largest multiple</strong> is chosen.</p>
+
+<p>Return <em>an integer denoting your account balance after making a purchase worth </em><code>purchaseAmount</code><em> dollars from the store.</em></p>
+
+<p><strong>Note:</strong> <code>0</code> is considered to be a multiple of <code>10</code> in this problem.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> purchaseAmount = 9
+<strong>Output:</strong> 90
+<strong>Explanation:</strong> In this example, the nearest multiple of 10 to 9 is 10. Hence, your account balance becomes 100 - 10 = 90.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> purchaseAmount = 15
+<strong>Output:</strong> 80
+<strong>Explanation:</strong> In this example, there are two nearest multiples of 10 to 15: 10 and 20. So, the larger multiple, 20, is chosen.
+Hence, your account balance becomes 100 - 20 = 80.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>0 &lt;= purchaseAmount &lt;= 100</code></li>
+</ul>
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+class Solution:
+    def accountBalanceAfterPurchase(self, purchaseAmount: int) -> int:
+        diff, x = 100, 0
+        for y in range(100, -1, -10):
+            if (t := abs(y - purchaseAmount)) < diff:
+                diff = t
+                x = y
+        return 100 - x
+```
+
+### **Java**
+
+```java
+class Solution {
+    public int accountBalanceAfterPurchase(int purchaseAmount) {
+        int diff = 100, x = 0;
+        for (int y = 100; y >= 0; y -= 10) {
+            int t = Math.abs(y - purchaseAmount);
+            if (t < diff) {
+                diff = t;
+                x = y;
+            }
+        }
+        return 100 - x;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int accountBalanceAfterPurchase(int purchaseAmount) {
+        int diff = 100, x = 0;
+        for (int y = 100; y >= 0; y -= 10) {
+            int t = abs(y - purchaseAmount);
+            if (t < diff) {
+                diff = t;
+                x = y;
+            }
+        }
+        return 100 - x;
+    }
+};
+```
+
+### **Go**
+
+```go
+func accountBalanceAfterPurchase(purchaseAmount int) int {
+	diff, x := 100, 0
+	for y := 100; y >= 0; y -= 10 {
+		t := abs(y - purchaseAmount)
+		if t < diff {
+			diff = t
+			x = y
+		}
+	}
+	return 100 - x
+}
+
+func abs(x int) int {
+	if x < 0 {
+		return -x
+	}
+	return x
+}
+```
+
+### **TypeScript**
+
+```ts
+function accountBalanceAfterPurchase(purchaseAmount: number): number {
+    let [diff, x] = [100, 0];
+    for (let y = 100; y >= 0; y -= 10) {
+        const t = Math.abs(y - purchaseAmount);
+        if (t < diff) {
+            diff = t;
+            x = y;
+        }
+    }
+    return 100 - x;
+}
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2800-2899/2806.Account Balance After Rounded Purchase/Solution.cpp

@@ -0,0 +1,14 @@
+class Solution {
+public:
+    int accountBalanceAfterPurchase(int purchaseAmount) {
+        int diff = 100, x = 0;
+        for (int y = 100; y >= 0; y -= 10) {
+            int t = abs(y - purchaseAmount);
+            if (t < diff) {
+                diff = t;
+                x = y;
+            }
+        }
+        return 100 - x;
+    }
+};

solution/2800-2899/2806.Account Balance After Rounded Purchase/Solution.go

@@ -0,0 +1,18 @@
+func accountBalanceAfterPurchase(purchaseAmount int) int {
+	diff, x := 100, 0
+	for y := 100; y >= 0; y -= 10 {
+		t := abs(y - purchaseAmount)
+		if t < diff {
+			diff = t
+			x = y
+		}
+	}
+	return 100 - x
+}
+
+func abs(x int) int {
+	if x < 0 {
+		return -x
+	}
+	return x
+}

solution/2800-2899/2806.Account Balance After Rounded Purchase/Solution.java

@@ -0,0 +1,13 @@
+class Solution {
+    public int accountBalanceAfterPurchase(int purchaseAmount) {
+        int diff = 100, x = 0;
+        for (int y = 100; y >= 0; y -= 10) {
+            int t = Math.abs(y - purchaseAmount);
+            if (t < diff) {
+                diff = t;
+                x = y;
+            }
+        }
+        return 100 - x;
+    }
+}

solution/2800-2899/2806.Account Balance After Rounded Purchase/Solution.py

@@ -0,0 +1,8 @@
+class Solution:
+    def accountBalanceAfterPurchase(self, purchaseAmount: int) -> int:
+        diff, x = 100, 0
+        for y in range(100, -1, -10):
+            if (t := abs(y - purchaseAmount)) < diff:
+                diff = t
+                x = y
+        return 100 - x

solution/2800-2899/2806.Account Balance After Rounded Purchase/Solution.ts

@@ -0,0 +1,11 @@
+function accountBalanceAfterPurchase(purchaseAmount: number): number {
+    let [diff, x] = [100, 0];
+    for (let y = 100; y >= 0; y -= 10) {
+        const t = Math.abs(y - purchaseAmount);
+        if (t < diff) {
+            diff = t;
+            x = y;
+        }
+    }
+    return 100 - x;
+}