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feat: add solutions to lc problems: No.2591~2598
* No.2591.Distribute Money to Maximum Children * No.2592.Maximize Greatness of an Array * No.2593.Find Score of an Array After Marking All Elements * No.2594.Minimum Time to Repair Cars * No.2595.Number of Even and Odd Bits * No.2596.Check Knight Tour Configuration * No.2597.The Number of Beautiful Subsets * No.2598.Smallest Missing Non-negative Integer After Operations
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‎solution/2800-2899/2890.Design a Todo List/README.md ‎solution/2500-2599/2590.Design a Todo List/README.md

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# [2890. Design a Todo List](https://leetcode.cn/problems/design-a-todo-list)
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# [2590. Design a Todo List](https://leetcode.cn/problems/design-a-todo-list)
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[English Version](/solution/2800-2899/2890.Design%20a%20Todo%20List/README_EN.md)
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[English Version](/solution/2500-2599/2590.Design%20a%20Todo%20List/README_EN.md)
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## 题目描述
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‎solution/2800-2899/2890.Design a Todo List/README_EN.md ‎solution/2500-2599/2590.Design a Todo List/README_EN.md

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# [2890. Design a Todo List](https://leetcode.com/problems/design-a-todo-list)
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# [2590. Design a Todo List](https://leetcode.com/problems/design-a-todo-list)
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[中文文档](/solution/2800-2899/2890.Design%20a%20Todo%20List/README.md)
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[中文文档](/solution/2500-2599/2590.Design%20a%20Todo%20List/README.md)
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## Description
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‎solution/2800-2899/2890.Design a Todo List/Solution.java ‎solution/2500-2599/2590.Design a Todo List/Solution.java

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‎solution/2800-2899/2890.Design a Todo List/Solution.py ‎solution/2500-2599/2590.Design a Todo List/Solution.py

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‎solution/2500-2599/2591.Distribute Money to Maximum Children/README.md

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# [2591. 将钱分给最多的儿童](https://leetcode.cn/problems/distribute-money-to-maximum-children)
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[English Version](/solution/2500-2599/2591.Distribute%20Money%20to%20Maximum%20Children/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>给你一个整数&nbsp;<code>money</code>&nbsp;，表示你总共有的钱数（单位为美元）和另一个整数&nbsp;<code>children</code>&nbsp;，表示你要将钱分配给多少个儿童。</p>
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<p>你需要按照如下规则分配：</p>
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<ul>
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<li>所有的钱都必须被分配。</li>
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<li>每个儿童至少获得&nbsp;<code>1</code>&nbsp;美元。</li>
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<li>没有人获得 <code>4</code>&nbsp;美元。</li>
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</ul>
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<p>请你按照上述规则分配金钱，并返回 <strong>最多</strong>&nbsp;有多少个儿童获得 <strong>恰好</strong><em>&nbsp;</em><code>8</code>&nbsp;美元。如果没有任何分配方案，返回&nbsp;<code>-1</code>&nbsp;。</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>money = 20, children = 3
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<b>输出：</b>1
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<b>解释：</b>
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最多获得 8 美元的儿童数为 1 。一种分配方案为：
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- 给第一个儿童分配 8 美元。
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- 给第二个儿童分配 9 美元。
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- 给第三个儿童分配 3 美元。
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没有分配方案能让获得 8 美元的儿童数超过 1 。
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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>money = 16, children = 2
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<b>输出：</b>2
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<b>解释：</b>每个儿童都可以获得 8 美元。
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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>1 &lt;= money &lt;= 200</code></li>
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<li><code>2 &lt;= children &lt;= 30</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：分类讨论**
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如果 $money \lt children$，那么一定存在儿童没有分到钱，返回 $-1$。
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如果 $money \gt 8 \times children$，那么有 $children-1$ 个儿童获得了 $8$ 美元，剩下的一个儿童获得了 $money - 8 \times (children-1)$ 美元，返回 $children-1$。
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如果 $money = 8 \times children - 4$，那么有 $children-2$ 个儿童获得了 $8$ 美元，剩下的两个儿童分摊剩下的 $12$ 美元（只要不是 $4$, $8$ 美元就行），返回 $children-2$。
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如果，我们假设有 $x$ 个儿童获得了 $8$ 美元，那么剩下的钱为 $money- 8 \times x$，只要保证大于等于剩下的儿童数 $children-x$，就可以满足题意。因此，我们只需要求出 $x$ 的最大值，即为答案。
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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 distMoney(self, money: int, children: int) -> int:
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if money < children:
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return -1
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if money > 8 * children:
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return children - 1
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if money == 8 * children - 4:
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return children - 2
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# money-8x >= children-x, x <= (money-children)/7
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return (money - children) // 7
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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 distMoney(int money, int children) {
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if (money < children) {
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return -1;
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}
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if (money > 8 * children) {
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return children - 1;
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}
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if (money == 8 * children - 4) {
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return children - 2;
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}
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// money-8x >= children-x, x <= (money-children)/7
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return (money - children) / 7;
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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 distMoney(int money, int children) {
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if (money < children) {
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return -1;
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}
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if (money > 8 * children) {
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return children - 1;
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}
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if (money == 8 * children - 4) {
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return children - 2;
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}
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// money-8x >= children-x, x <= (money-children)/7
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return (money - children) / 7;
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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 distMoney(money int, children int) int {
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if money < children {
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return -1
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}
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if money > 8*children {
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return children - 1
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}
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if money == 8*children-4 {
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return children - 2
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}
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// money-8x >= children-x, x <= (money-children)/7
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return (money - children) / 7
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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/2500-2599/2591.Distribute Money to Maximum Children/README_EN.md

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# [2591. Distribute Money to Maximum Children](https://leetcode.com/problems/distribute-money-to-maximum-children)
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[中文文档](/solution/2500-2599/2591.Distribute%20Money%20to%20Maximum%20Children/README.md)
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## Description
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<p>You are given an integer <code>money</code> denoting the amount of money (in dollars) that you have and another integer <code>children</code> denoting the number of children that you must distribute the money to.</p>
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<p>You have to distribute the money according to the following rules:</p>
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<ul>
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<li>All money must be distributed.</li>
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<li>Everyone must receive at least <code>1</code> dollar.</li>
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<li>Nobody receives <code>4</code> dollars.</li>
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</ul>
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<p>Return <em>the <strong>maximum</strong> number of children who may receive <strong>exactly</strong> </em><code>8</code> <em>dollars if you distribute the money according to the aforementioned rules</em>. If there is no way to distribute the money, return <code>-1</code>.</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> money = 20, children = 3
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<strong>Output:</strong> 1
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<strong>Explanation:</strong>
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The maximum number of children with 8 dollars will be 1. One of the ways to distribute the money is:
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- 8 dollars to the first child.
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- 9 dollars to the second child.
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- 3 dollars to the third child.
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It can be proven that no distribution exists such that number of children getting 8 dollars is greater than 1.
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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> money = 16, children = 2
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<strong>Output:</strong> 2
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<strong>Explanation:</strong> Each child can be given 8 dollars.
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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>1 &lt;= money &lt;= 200</code></li>
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<li><code>2 &lt;= children &lt;= 30</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 distMoney(self, money: int, children: int) -> int:
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if money < children:
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return -1
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if money > 8 * children:
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return children - 1
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if money == 8 * children - 4:
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return children - 2
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# money-8x >= children-x, x <= (money-children)/7
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return (money - children) // 7
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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 distMoney(int money, int children) {
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if (money < children) {
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return -1;
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}
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if (money > 8 * children) {
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return children - 1;
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}
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if (money == 8 * children - 4) {
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return children - 2;
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}
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// money-8x >= children-x, x <= (money-children)/7
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return (money - children) / 7;
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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 distMoney(int money, int children) {
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if (money < children) {
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return -1;
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}
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if (money > 8 * children) {
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return children - 1;
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}
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if (money == 8 * children - 4) {
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return children - 2;
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}
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// money-8x >= children-x, x <= (money-children)/7
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return (money - children) / 7;
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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 distMoney(money int, children int) int {
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if money < children {
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return -1
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}
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if money > 8*children {
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return children - 1
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}
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if money == 8*children-4 {
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return children - 2
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}
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// money-8x >= children-x, x <= (money-children)/7
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return (money - children) / 7
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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/2500-2599/2591.Distribute Money to Maximum Children/Solution.cpp

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class Solution {
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public:
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int distMoney(int money, int children) {
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if (money < children) {
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return -1;
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}
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if (money > 8 * children) {
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return children - 1;
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}
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if (money == 8 * children - 4) {
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return children - 2;
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}
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// money-8x >= children-x, x <= (money-children)/7
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return (money - children) / 7;
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}
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};

‎solution/2500-2599/2591.Distribute Money to Maximum Children/Solution.go

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func distMoney(money int, children int) int {
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if money < children {
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return -1
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}
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if money > 8*children {
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return children - 1
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}
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if money == 8*children-4 {
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return children - 2
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}
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// money-8x >= children-x, x <= (money-children)/7
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return (money - children) / 7
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}

‎solution/2500-2599/2591.Distribute Money to Maximum Children/Solution.java

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class Solution {
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public int distMoney(int money, int children) {
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if (money < children) {
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return -1;
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}
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if (money > 8 * children) {
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return children - 1;
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}
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if (money == 8 * children - 4) {
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return children - 2;
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}
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// money-8x >= children-x, x <= (money-children)/7
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return (money - children) / 7;
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}
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}

‎solution/2500-2599/2591.Distribute Money to Maximum Children/Solution.py

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class Solution:
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def distMoney(self, money: int, children: int) -> int:
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if money < children:
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return -1
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if money > 8 * children:
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return children - 1
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if money == 8 * children - 4:
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return children - 2
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# money-8x >= children-x, x <= (money-children)/7
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return (money - children) // 7

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