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feat: add solutions to lc problems: No.2806~2813 (#1407)
* No.2806.Account Balance After Rounded Purchase * No.2807.Insert Greatest Common Divisors in Linked List * No.2808.Minimum Seconds to Equalize a Circular Array * No.2809.Minimum Time to Make Array Sum At Most x * No.2810.Faulty Keyboard * No.2811.Check if it is Possible to Split Array * No.2812.Find the Safest Path in a Grid * No.2813.Maximum Elegance of a K-Length Subsequence
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‎solution/2800-2899/2806.Account Balance After Rounded Purchase/README.md

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# [2806. 取整购买后的账户余额](https://leetcode.cn/problems/account-balance-after-rounded-purchase)
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[English Version](/solution/2800-2899/2806.Account%20Balance%20After%20Rounded%20Purchase/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>一开始，你的银行账户里有&nbsp;<code>100</code>&nbsp;块钱。</p>
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<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>
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<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>
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<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>
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<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>
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<p><strong>注意：</strong> <code>0</code>&nbsp;也是&nbsp;<code>10</code>&nbsp;的倍数。</p>
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<p>&nbsp;</p>
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<p><strong>示例 1：</strong></p>
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<pre><b>输入：</b>purchaseAmount = 9
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<b>输出：</b>90
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<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 。
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</pre>
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<p><strong>示例 2：</strong></p>
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<pre><b>输入：</b>purchaseAmount = 15
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<b>输出：</b>80
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<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。
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所以你的账户余额为 100 - 20 = 80 。
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</pre>
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<p>&nbsp;</p>
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<p><strong>提示：</strong></p>
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<ul>
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<li><code>0 &lt;= purchaseAmount &lt;= 100</code></li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：枚举 + 模拟**
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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 $100 - x$。
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时间复杂度 $O(1)$，空间复杂度 $O(1)$。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def accountBalanceAfterPurchase(self, purchaseAmount: int) -> int:
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diff, x = 100, 0
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for y in range(100, -1, -10):
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if (t := abs(y - purchaseAmount)) < diff:
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diff = t
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x = y
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return 100 - x
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int accountBalanceAfterPurchase(int purchaseAmount) {
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int diff = 100, x = 0;
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for (int y = 100; y >= 0; y -= 10) {
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int t = Math.abs(y - purchaseAmount);
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if (t < diff) {
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diff = t;
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x = y;
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}
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}
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return 100 - x;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int accountBalanceAfterPurchase(int purchaseAmount) {
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int diff = 100, x = 0;
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for (int y = 100; y >= 0; y -= 10) {
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int t = abs(y - purchaseAmount);
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if (t < diff) {
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diff = t;
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x = y;
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}
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}
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return 100 - x;
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}
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};
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```
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### **Go**
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```go
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func accountBalanceAfterPurchase(purchaseAmount int) int {
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diff, x := 100, 0
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for y := 100; y >= 0; y -= 10 {
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t := abs(y - purchaseAmount)
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if t < diff {
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diff = t
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x = y
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}
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}
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return 100 - x
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}
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func abs(x int) int {
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if x < 0 {
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return -x
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}
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return x
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}
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```
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### **TypeScript**
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```ts
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function accountBalanceAfterPurchase(purchaseAmount: number): number {
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let [diff, x] = [100, 0];
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for (let y = 100; y >= 0; y -= 10) {
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const t = Math.abs(y - purchaseAmount);
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if (t < diff) {
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diff = t;
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x = y;
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}
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}
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return 100 - x;
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}
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```
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### **...**
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```
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```
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<!-- tabs:end -->

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

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

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

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

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

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

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

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

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

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

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

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

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