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feat: add new lc problems: No.2894~2897 (doocs#1764)
* No.2894.Divisible and Non-divisible Sums Difference * No.2895.Minimum Processing Time * No.2896.Apply Operations to Make Two Strings Equal * No.2897.Apply Operations on Array to Maximize Sum of Squares
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# [2894. 分类求和并作差](https://leetcode.cn/problems/divisible-and-non-divisible-sums-difference)
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[English Version](/solution/2800-2899/2894.Divisible%20and%20Non-divisible%20Sums%20Difference/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>给你两个正整数 <code>n</code> 和 <code>m</code> 。</p>
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<p>现定义两个整数 <code>num1</code> 和 <code>num2</code> ,如下所示:</p>
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<ul>
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<li><code>num1</code>:范围 <code>[1, n]</code> 内所有 <strong>无法被 </strong><code>m</code><strong> 整除</strong> 的整数之和。</li>
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<li><code>num2</code>:范围 <code>[1, n]</code> 内所有 <strong>能够被 </strong><code>m</code><strong> 整除</strong> 的整数之和。</li>
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</ul>
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<p>返回整数 <code>num1 - num2</code> 。</p>
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<p>&nbsp;</p>
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<p><strong class="example">示例 1:</strong></p>
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<pre>
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<strong>输入:</strong>n = 10, m = 3
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<strong>输出:</strong>19
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<strong>解释:</strong>在这个示例中:
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- 范围 [1, 10] 内无法被 3 整除的整数为 [1,2,4,5,7,8,10] ,num1 = 这些整数之和 = 37 。
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- 范围 [1, 10] 内能够被 3 整除的整数为 [3,6,9] ,num2 = 这些整数之和 = 18 。
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返回 37 - 18 = 19 作为答案。
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</pre>
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<p><strong class="example">示例 2:</strong></p>
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<pre>
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<strong>输入:</strong>n = 5, m = 6
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<strong>输出:</strong>15
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<strong>解释:</strong>在这个示例中:
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- 范围 [1, 5] 内无法被 6 整除的整数为 [1,2,3,4,5] ,num1 = 这些整数之和 = 15 。
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- 范围 [1, 5] 内能够被 6 整除的整数为 [] ,num2 = 这些整数之和 = 0 。
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返回 15 - 0 = 15 作为答案。
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</pre>
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<p><strong class="example">示例 3:</strong></p>
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<pre>
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<strong>输入:</strong>n = 5, m = 1
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<strong>输出:</strong>-15
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<strong>解释:</strong>在这个示例中:
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- 范围 [1, 5] 内无法被 1 整除的整数为 [] ,num1 = 这些整数之和 = 0 。
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- 范围 [1, 5] 内能够被 1 整除的整数为 [1,2,3,4,5] ,num2 = 这些整数之和 = 15 。
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返回 0 - 15 = -15 作为答案。
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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;= n, m &lt;= 1000</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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<!-- 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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```
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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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```
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### **C++**
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```cpp
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```
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### **Go**
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```go
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```
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### **...**
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```
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```
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<!-- tabs:end -->
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# [2894. Divisible and Non-divisible Sums Difference](https://leetcode.com/problems/divisible-and-non-divisible-sums-difference)
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[中文文档](/solution/2800-2899/2894.Divisible%20and%20Non-divisible%20Sums%20Difference/README.md)
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## Description
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<p>You are given positive integers <code>n</code> and <code>m</code>.</p>
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<p>Define two integers, <code>num1</code> and <code>num2</code>, as follows:</p>
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<ul>
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<li><code>num1</code>: The sum of all integers in the range <code>[1, n]</code> that are <strong>not divisible</strong> by <code>m</code>.</li>
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<li><code>num2</code>: The sum of all integers in the range <code>[1, n]</code> that are <strong>divisible</strong> by <code>m</code>.</li>
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</ul>
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<p>Return <em>the integer</em> <code>num1 - num2</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> n = 10, m = 3
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<strong>Output:</strong> 19
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<strong>Explanation:</strong> In the given example:
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- Integers in the range [1, 10] that are not divisible by 3 are [1,2,4,5,7,8,10], num1 is the sum of those integers = 37.
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- Integers in the range [1, 10] that are divisible by 3 are [3,6,9], num2 is the sum of those integers = 18.
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We return 37 - 18 = 19 as the answer.
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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> n = 5, m = 6
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<strong>Output:</strong> 15
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<strong>Explanation:</strong> In the given example:
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- Integers in the range [1, 5] that are not divisible by 6 are [1,2,3,4,5], num1 is the sum of those integers = 15.
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- Integers in the range [1, 5] that are divisible by 6 are [], num2 is the sum of those integers = 0.
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We return 15 - 0 = 15 as the answer.
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</pre>
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<p><strong class="example">Example 3:</strong></p>
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<pre>
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<strong>Input:</strong> n = 5, m = 1
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<strong>Output:</strong> -15
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<strong>Explanation:</strong> In the given example:
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- Integers in the range [1, 5] that are not divisible by 1 are [], num1 is the sum of those integers = 0.
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- Integers in the range [1, 5] that are divisible by 1 are [1,2,3,4,5], num2 is the sum of those integers = 15.
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We return 0 - 15 = -15 as the answer.
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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;= n, m &lt;= 1000</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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```
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### **Java**
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```java
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```
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### **C++**
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```cpp
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```
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### **Go**
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```go
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```
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### **...**
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```
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```
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# [2895. 最小处理时间](https://leetcode.cn/problems/minimum-processing-time)
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[English Version](/solution/2800-2899/2895.Minimum%20Processing%20Time/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>你有 <code>n</code> 颗处理器,每颗处理器都有 <code>4</code> 个核心。现有 <code>n * 4</code> 个待执行任务,每个核心只执行 <strong>一个</strong> 任务。</p>
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<p>给你一个下标从 <strong>0</strong> 开始的整数数组 <code>processorTime</code> ,表示每颗处理器最早空闲时间。另给你一个下标从 <strong>0</strong> 开始的整数数组 <code>tasks</code> ,表示执行每个任务所需的时间。返回所有任务都执行完毕需要的 <strong>最小时间</strong> 。</p>
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<p>注意:每个核心独立执行任务。</p>
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<p>&nbsp;</p>
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<p><strong>示例 1:</strong></p>
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<pre>
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<strong>输入:</strong>processorTime = [8,10], tasks = [2,2,3,1,8,7,4,5]
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<strong>输出:</strong>16
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<strong>解释:</strong>
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最优的方案是将下标为 4, 5, 6, 7 的任务分配给第一颗处理器(最早空闲时间 time = 8),下标为 0, 1, 2, 3 的任务分配给第二颗处理器(最早空闲时间 time = 10)。
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第一颗处理器执行完所有任务需要花费的时间 = max(8 + 8, 8 + 7, 8 + 4, 8 + 5) = 16 。
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第二颗处理器执行完所有任务需要花费的时间 = max(10 + 2, 10 + 2, 10 + 3, 10 + 1) = 13 。
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因此,可以证明执行完所有任务需要花费的最小时间是 16 。</pre>
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<p><strong>示例 2:</strong></p>
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<pre>
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<strong>输入:</strong>processorTime = [10,20], tasks = [2,3,1,2,5,8,4,3]
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<strong>输出:</strong>23
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<strong>解释:</strong>
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最优的方案是将下标为 1, 4, 5, 6 的任务分配给第一颗处理器(最早空闲时间 time = 10),下标为 0, 2, 3, 7 的任务分配给第二颗处理器(最早空闲时间 time = 20)。
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第一颗处理器执行完所有任务需要花费的时间 = max(10 + 3, 10 + 5, 10 + 8, 10 + 4) = 18 。
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第二颗处理器执行完所有任务需要花费的时间 = max(20 + 2, 20 + 1, 20 + 2, 20 + 3) = 23 。
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因此,可以证明执行完所有任务需要花费的最小时间是 23 。
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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;= n == processorTime.length &lt;= 25000</code></li>
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<li><code>1 &lt;= tasks.length &lt;= 10<sup>5</sup></code></li>
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<li><code>0 &lt;= processorTime[i] &lt;= 10<sup>9</sup></code></li>
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<li><code>1 &lt;= tasks[i] &lt;= 10<sup>9</sup></code></li>
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<li><code>tasks.length == 4 * n</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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<!-- 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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```
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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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```
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### **C++**
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```cpp
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```
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### **Go**
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```go
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```
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### **...**
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```
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```
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<!-- tabs:end -->
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# [2895. Minimum Processing Time](https://leetcode.com/problems/minimum-processing-time)
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[中文文档](/solution/2800-2899/2895.Minimum%20Processing%20Time/README.md)
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## Description
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<p>You have <code>n</code> processors each having <code>4</code> cores and <code>n * 4</code> tasks that need to be executed such that each core should perform only <strong>one</strong> task.</p>
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<p>Given a <strong>0-indexed</strong> integer array <code>processorTime</code> representing the time at which each processor becomes available for the first time and a <strong>0-indexed </strong>integer array <code>tasks</code> representing the time it takes to execute each task, return <em>the <strong>minimum</strong> time when all of the tasks have been executed by the processors.</em></p>
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<p><strong>Note: </strong>Each core executes the task independently of the others.</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> processorTime = [8,10], tasks = [2,2,3,1,8,7,4,5]
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<strong>Output:</strong> 16
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<strong>Explanation:</strong>
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It&#39;s optimal to assign the tasks at indexes 4, 5, 6, 7 to the first processor which becomes available at time = 8, and the tasks at indexes 0, 1, 2, 3 to the second processor which becomes available at time = 10.
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Time taken by the first processor to finish execution of all tasks = max(8 + 8, 8 + 7, 8 + 4, 8 + 5) = 16.
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Time taken by the second processor to finish execution of all tasks = max(10 + 2, 10 + 2, 10 + 3, 10 + 1) = 13.
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Hence, it can be shown that the minimum time taken to execute all the tasks is 16.</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> processorTime = [10,20], tasks = [2,3,1,2,5,8,4,3]
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<strong>Output:</strong> 23
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<strong>Explanation:</strong>
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It&#39;s optimal to assign the tasks at indexes 1, 4, 5, 6 to the first processor which becomes available at time = 10, and the tasks at indexes 0, 2, 3, 7 to the second processor which becomes available at time = 20.
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Time taken by the first processor to finish execution of all tasks = max(10 + 3, 10 + 5, 10 + 8, 10 + 4) = 18.
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Time taken by the second processor to finish execution of all tasks = max(20 + 2, 20 + 1, 20 + 2, 20 + 3) = 23.
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Hence, it can be shown that the minimum time taken to execute all the tasks is 23.
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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;= n == processorTime.length &lt;= 25000</code></li>
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<li><code>1 &lt;= tasks.length &lt;= 10<sup>5</sup></code></li>
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<li><code>0 &lt;= processorTime[i] &lt;= 10<sup>9</sup></code></li>
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<li><code>1 &lt;= tasks[i] &lt;= 10<sup>9</sup></code></li>
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<li><code>tasks.length == 4 * n</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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```
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### **Java**
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```java
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```
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### **C++**
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```cpp
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```
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### **Go**
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```go
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```
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### **...**
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```
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```
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<!-- tabs:end -->

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