chore: sync lc problems No.4004, No.4005

solution/4000-4099/4004.Minimum Moves to Balance Circular Array II/README.md

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+---
+comments: true
+difficulty: 困难
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/4000-4099/4004.Minimum%20Moves%20to%20Balance%20Circular%20Array%20II/README.md
+---
+
+<!-- problem:start -->
+
+# [4004. 使循环数组余额非负的最少移动次数 II 🔒](https://leetcode.cn/problems/minimum-moves-to-balance-circular-array-ii)
+
+[English Version](/solution/4000-4099/4004.Minimum%20Moves%20to%20Balance%20Circular%20Array%20II/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>给定一个长度为 <code>n</code> 的 <span data-keyword="circular-array">环形数组</span> <code>balance</code>，其中 <code>balance[i]</code>&nbsp;是第 <code>i</code> 个人的净余额。</p>
+
+<p>在一次操作中，一个人可以向其左侧或右侧的相邻人员转移&nbsp;<strong>恰好</strong> 1 单位的余额。</p>
+
+<p>返回使每个人的余额都变为&nbsp;<strong>非负&nbsp;</strong>所需的&nbsp;<strong>最少&nbsp;</strong>操作次数。如果无法做到，则返回 -1。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong><span class="example-io">balance = [-1,2,-1]</span></p>
+
+<p><strong>输出：</strong><span class="example-io">2</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>一种最优的操作序列如下：</p>
+
+<ul>
+	<li>从 <code>i = 1</code> 向 <code>i = 0</code> 转移 1 单位余额，得到 <code>balance = [0, 1, -1]</code></li>
+	<li>从 <code>i = 1</code> 向 <code>i = 2</code> 转移 1 单位余额，得到 <code>balance = [0, 0, 0]</code></li>
+</ul>
+
+<p>因此，所需的最少操作次数为 2。</p>
+</div>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong><span class="example-io">balance = [4,-1,-2]</span></p>
+
+<p><strong>输出：</strong><span class="example-io">3</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>一种最优的操作序列如下：</p>
+
+<ul>
+	<li>从 <code>i = 0</code> 向 <code>i = 1</code> 转移 1 单位余额，得到 <code>balance = [3, 0, -2]</code></li>
+	<li>从 <code>i = 0</code> 向 <code>i = 2</code> 转移 1 单位余额，得到 <code>balance = [2, 0, -1]</code></li>
+	<li>从 <code>i = 0</code> 向 <code>i = 2</code> 再转移 1 单位余额，得到 <code>balance = [1, 0, 0]</code></li>
+</ul>
+
+<p>因此，所需的最少操作次数为 3。</p>
+</div>
+
+<p><strong class="example">示例 3：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong><span class="example-io">balance = [-3,-3,5]</span></p>
+
+<p><strong>输出：</strong><span class="example-io">-1</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>对于 <code>balance = [-3, -3, 5]</code>，无法使所有人的余额都变为非负，因此答案为 -1。</p>
+</div>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n == balance.length &lt;= 1000</code></li>
+	<li><code>-10<sup>5</sup> &lt;= balance[i] &lt;= 10<sup>5</sup></code></li>
+</ul>
+
+<!-- description:end -->
+
+## 解法
+
+<!-- solution:start -->
+
+### 方法一
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+
+```
+
+#### Java
+
+```java
+
+```
+
+#### C++
+
+```cpp
+
+```
+
+#### Go
+
+```go
+
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->

solution/4000-4099/4004.Minimum Moves to Balance Circular Array II/README_EN.md

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+---
+comments: true
+difficulty: Hard
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/4000-4099/4004.Minimum%20Moves%20to%20Balance%20Circular%20Array%20II/README_EN.md
+---
+
+<!-- problem:start -->
+
+# [4004. Minimum Moves to Balance Circular Array II 🔒](https://leetcode.com/problems/minimum-moves-to-balance-circular-array-ii)
+
+[中文文档](/solution/4000-4099/4004.Minimum%20Moves%20to%20Balance%20Circular%20Array%20II/README.md)
+
+## Description
+
+<!-- description:start -->
+
+<p>You are given a <span data-keyword="circular-array">circular array</span> <code>balance</code> of length <code>n</code>, where <code>balance[i]</code> is the net balance of person <code>i</code>.</p>
+
+<p>In one move, a person can transfer <strong>exactly</strong> 1 unit of balance to either their left or right neighbor.</p>
+
+<p>Return the <strong>minimum</strong> number of moves required so that every person has a <strong>non-negative</strong> balance. If it is impossible, return -1.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">balance = [-1,2,-1]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">2</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>One optimal sequence of moves is:</p>
+
+<ul>
+	<li>Move 1 unit from <code>i = 1</code> to <code>i = 0</code>, resulting in <code>balance = [0, 1, -1]</code></li>
+	<li>Move 1 unit from <code>i = 1</code> to <code>i = 2</code>, resulting in <code>balance = [0, 0, 0]</code></li>
+</ul>
+
+<p>Thus, the minimum number of moves required is 2.</p>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">balance = [4,-1,-2]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">3</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>One optimal sequence of moves is:</p>
+
+<ul>
+	<li>Move 1 unit from <code>i = 0</code> to <code>i = 1</code>, resulting in <code>balance = [3, 0, -2]</code></li>
+	<li>Move 1 unit from <code>i = 0</code> to <code>i = 2</code>, resulting in <code>balance = [2, 0, -1]</code></li>
+	<li>Move 1 unit from <code>i = 0</code> to <code>i = 2</code>, resulting in <code>balance = [1, 0, 0]</code></li>
+</ul>
+
+<p>Thus, the minimum number of moves required is 3.</p>
+</div>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">balance = [-3,-3,5]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">-1</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>It is impossible to make all balances non-negative for <code>balance = [-3, -3, 5]</code>, so the answer is -1.</p>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n == balance.length &lt;= 1000</code></li>
+	<li><code>-10<sup>5</sup> &lt;= balance[i] &lt;= 10<sup>5</sup></code></li>
+</ul>
+
+<!-- description:end -->
+
+## Solutions
+
+<!-- solution:start -->
+
+### Solution 1
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+
+```
+
+#### Java
+
+```java
+
+```
+
+#### C++
+
+```cpp
+
+```
+
+#### Go
+
+```go
+
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->

solution/4000-4099/4005.Minimum Operations to Make Array Equal III/README.md

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+---
+comments: true
+difficulty: 困难
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/4000-4099/4005.Minimum%20Operations%20to%20Make%20Array%20Equal%20III/README.md
+---
+
+<!-- problem:start -->
+
+# [4005. Minimum Operations to Make Array Equal III 🔒](https://leetcode.cn/problems/minimum-operations-to-make-array-equal-iii)
+
+[English Version](/solution/4000-4099/4005.Minimum%20Operations%20to%20Make%20Array%20Equal%20III/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>You are given an integer array <code>nums</code>.</p>
+
+<p>In one operation, you may choose <strong>any</strong> element <code>nums[i]</code> and perform one of the following:</p>
+
+<ul>
+	<li><strong>Multiply</strong> <code>nums[i]</code> by an integer <code>k</code>, where <code>k &gt;= 2</code>.</li>
+	<li><strong>Divide</strong> <code>nums[i]</code> by an integer <code>k</code>, where <code>2 &lt;= k &lt; nums[i]</code>, provided that <code>nums[i]</code> is divisible by <code>k</code>.</li>
+</ul>
+
+<p>Return the <strong>minimum</strong> number of operations required to make all elements of <code>nums</code> <strong>equal</strong>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [6,12,8]</span></p>
+
+<p><strong>Output:</strong> 3</p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>We can perform following operates to make all numbers to 6:</p>
+
+<ul>
+	<li>Divide <code>nums[1] = 12</code> by 2 to get 6.</li>
+	<li>Divide <code>nums[2] = 8</code> by 4 to get 2.</li>
+	<li>Multiply <code>nums[2] = 2</code> by 3 to get 6.</li>
+</ul>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [5,15,20]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">2</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>We can perform following operates to make all numbers to 5:</p>
+
+<ul>
+	<li>Divide <code>nums[1] = 15</code> by 3 to get 5.</li>
+	<li>Divide <code>nums[2] = 20</code> by 4 to get 5.</li>
+</ul>
+</div>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [7,7,7]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">0</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>All elements are already equal, so no operations are needed.</p>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 10<sup>​​​​​​​9</sup></code></li>
+</ul>
+
+<!-- description:end -->
+
+## 解法
+
+<!-- solution:start -->
+
+### 方法一
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+
+```
+
+#### Java
+
+```java
+
+```
+
+#### C++
+
+```cpp
+
+```
+
+#### Go
+
+```go
+
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->