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‎solution/3600-3699/3652.Best Time to Buy and Sell Stock using Strategy/README.md

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---
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comments: true
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difficulty: 中等
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3600-3699/3652.Best%20Time%20to%20Buy%20and%20Sell%20Stock%20using%20Strategy/README.md
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<!-- problem:start -->
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# [3652. 按策略买卖股票的最佳时机](https://leetcode.cn/problems/best-time-to-buy-and-sell-stock-using-strategy)
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[English Version](/solution/3600-3699/3652.Best%20Time%20to%20Buy%20and%20Sell%20Stock%20using%20Strategy/README_EN.md)
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## 题目描述
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<!-- description:start -->
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<p>给你两个整数数组 <code>prices</code> 和 <code>strategy</code>，其中：</p>
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<ul>
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<li><code>prices[i]</code> 表示第 <code>i</code> 天某股票的价格。</li>
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<li><code>strategy[i]</code> 表示第 <code>i</code> 天的交易策略，其中：
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<ul>
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<li><code>-1</code> 表示买入一单位股票。</li>
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<li><code>0</code> 表示持有股票。</li>
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<li><code>1</code> 表示卖出一单位股票。</li>
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</ul>
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</li>
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</ul>
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<p>同时给你一个&nbsp;<strong>偶数&nbsp;</strong>整数 <code>k</code>，你可以对 <code>strategy</code> 进行&nbsp;<strong>最多一次&nbsp;</strong>修改。一次修改包括：</p>
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<ul>
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<li>选择 <code>strategy</code> 中恰好 <code>k</code> 个&nbsp;<strong>连续&nbsp;</strong>元素。</li>
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<li>将前 <code>k / 2</code> 个元素设为 <code>0</code>（持有）。</li>
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<li>将后 <code>k / 2</code> 个元素设为 <code>1</code>（卖出）。</li>
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</ul>
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<p><strong>利润&nbsp;</strong>定义为所有天数中 <code>strategy[i] * prices[i]</code> 的&nbsp;<strong>总和&nbsp;</strong>。</p>
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<p>返回你可以获得的&nbsp;<strong>最大&nbsp;</strong>可能利润。</p>
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<p><strong>注意：</strong> 没有预算或股票持有数量的限制，因此所有买入和卖出操作均可行，无需考虑过去的操作。</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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<div class="example-block">
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<p><strong>输入：</strong> <span class="example-io">prices = [4,2,8], strategy = [-1,0,1], k = 2</span></p>
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<p><strong>输出：</strong> <span class="example-io">10</span></p>
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<p><strong>解释：</strong></p>
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<table style="border: 1px solid black;">
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<thead>
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<tr>
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<th style="border: 1px solid black;">修改</th>
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<th style="border: 1px solid black;">策略</th>
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<th style="border: 1px solid black;">利润计算</th>
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<th style="border: 1px solid black;">利润</th>
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</tr>
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</thead>
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<tbody>
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<tr>
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<td style="border: 1px solid black;">原始</td>
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<td style="border: 1px solid black;">[-1, 0, 1]</td>
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<td style="border: 1px solid black;">(-1 × 4) + (0 × 2) + (1 × 8) = -4 + 0 + 8</td>
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<td style="border: 1px solid black;">4</td>
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</tr>
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<tr>
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<td style="border: 1px solid black;">修改 [0, 1]</td>
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<td style="border: 1px solid black;">[0, 1, 1]</td>
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<td style="border: 1px solid black;">(0 × 4) + (1 × 2) + (1 × 8) = 0 + 2 + 8</td>
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<td style="border: 1px solid black;">10</td>
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</tr>
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<tr>
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<td style="border: 1px solid black;">修改 [1, 2]</td>
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<td style="border: 1px solid black;">[-1, 0, 1]</td>
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<td style="border: 1px solid black;">(-1 × 4) + (0 × 2) + (1 × 8) = -4 + 0 + 8</td>
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<td style="border: 1px solid black;">4</td>
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</tr>
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</tbody>
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</table>
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<p>因此，最大可能利润是 10，通过修改子数组 <code>[0, 1]</code> 实现。</p>
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</div>
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<p><strong class="example">示例 2：</strong></p>
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<div class="example-block">
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<p><strong>输入：</strong> <span class="example-io">prices = [5,4,3], strategy = [1,1,0], k = 2</span></p>
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<p><strong>输出：</strong> <span class="example-io">9</span></p>
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<p><strong>解释：</strong></p>
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<div class="example-block">
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<table style="border: 1px solid black;">
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<thead>
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<tr>
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<th style="border: 1px solid black;">修改</th>
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<th style="border: 1px solid black;">策略</th>
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<th style="border: 1px solid black;">利润计算</th>
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<th style="border: 1px solid black;">利润</th>
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</tr>
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</thead>
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<tbody>
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<tr>
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<td style="border: 1px solid black;">原始</td>
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<td style="border: 1px solid black;">[1, 1, 0]</td>
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<td style="border: 1px solid black;">(1 × 5) + (1 × 4) + (0 × 3) = 5 + 4 + 0</td>
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<td style="border: 1px solid black;">9</td>
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</tr>
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<tr>
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<td style="border: 1px solid black;">修改 [0, 1]</td>
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<td style="border: 1px solid black;">[0, 1, 0]</td>
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<td style="border: 1px solid black;">(0 × 5) + (1 × 4) + (0 × 3) = 0 + 4 + 0</td>
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<td style="border: 1px solid black;">4</td>
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</tr>
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<tr>
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<td style="border: 1px solid black;">修改 [1, 2]</td>
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<td style="border: 1px solid black;">[1, 0, 1]</td>
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<td style="border: 1px solid black;">(1 × 5) + (0 × 4) + (1 × 3) = 5 + 0 + 3</td>
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<td style="border: 1px solid black;">8</td>
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</tr>
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</tbody>
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</table>
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<p>因此，最大可能利润是 9，无需任何修改即可达成。</p>
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</div>
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</div>
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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>2 &lt;= prices.length == strategy.length &lt;= 10<sup>5</sup></code></li>
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<li><code>1 &lt;= prices[i] &lt;= 10<sup>5</sup></code></li>
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<li><code>-1 &lt;= strategy[i] &lt;= 1</code></li>
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<li><code>2 &lt;= k &lt;= prices.length</code></li>
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<li><code>k</code> 是偶数</li>
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</ul>
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<!-- description:end -->
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## 解法
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<!-- solution:start -->
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### 方法一
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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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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->

‎solution/3600-3699/3652.Best Time to Buy and Sell Stock using Strategy/README_EN.md

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comments: true
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difficulty: Medium
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3600-3699/3652.Best%20Time%20to%20Buy%20and%20Sell%20Stock%20using%20Strategy/README_EN.md
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<!-- problem:start -->
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# [3652. Best Time to Buy and Sell Stock using Strategy](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-using-strategy)
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[中文文档](/solution/3600-3699/3652.Best%20Time%20to%20Buy%20and%20Sell%20Stock%20using%20Strategy/README.md)
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## Description
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<!-- description:start -->
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<p>You are given two integer arrays <code>prices</code> and <code>strategy</code>, where:</p>
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<ul>
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<li><code>prices[i]</code> is the price of a given stock on the <code>i<sup>th</sup></code> day.</li>
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<li><code>strategy[i]</code> represents a trading action on the <code>i<sup>th</sup></code> day, where:
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<ul>
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<li><code>-1</code> indicates buying one unit of the stock.</li>
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<li><code>0</code> indicates holding the stock.</li>
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<li><code>1</code> indicates selling one unit of the stock.</li>
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</ul>
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</li>
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</ul>
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<p>You are also given an <strong>even</strong> integer <code>k</code>, and may perform <strong>at most one</strong> modification to <code>strategy</code>. A modification consists of:</p>
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<ul>
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<li>Selecting exactly <code>k</code> <strong>consecutive</strong> elements in <code>strategy</code>.</li>
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<li>Set the <strong>first</strong> <code>k / 2</code> elements to <code>0</code> (hold).</li>
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<li>Set the <strong>last</strong> <code>k / 2</code> elements to <code>1</code> (sell).</li>
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</ul>
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<p>The <strong>profit</strong> is defined as the <strong>sum</strong> of <code>strategy[i] * prices[i]</code> across all days.</p>
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<p>Return the <strong>maximum</strong> possible profit you can achieve.</p>
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<p><strong>Note:</strong> There are no constraints on budget or stock ownership, so all buy and sell operations are feasible regardless of past actions.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<div class="example-block">
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<p><strong>Input:</strong> <span class="example-io">prices = [4,2,8], strategy = [-1,0,1], k = 2</span></p>
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<p><strong>Output:</strong> <span class="example-io">10</span></p>
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<p><strong>Explanation:</strong></p>
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<table style="border: 1px solid black;">
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<thead>
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<tr>
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<th style="border: 1px solid black;">Modification</th>
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<th style="border: 1px solid black;">Strategy</th>
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<th style="border: 1px solid black;">Profit Calculation</th>
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<th style="border: 1px solid black;">Profit</th>
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</tr>
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</thead>
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<tbody>
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<tr>
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<td style="border: 1px solid black;">Original</td>
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<td style="border: 1px solid black;">[-1, 0, 1]</td>
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<td style="border: 1px solid black;">(-1 &times; 4) + (0 &times; 2) + (1 &times; 8) = -4 + 0 + 8</td>
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<td style="border: 1px solid black;">4</td>
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</tr>
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<tr>
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<td style="border: 1px solid black;">Modify [0, 1]</td>
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<td style="border: 1px solid black;">[0, 1, 1]</td>
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<td style="border: 1px solid black;">(0 &times; 4) + (1 &times; 2) + (1 &times; 8) = 0 + 2 + 8</td>
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<td style="border: 1px solid black;">10</td>
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</tr>
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<tr>
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<td style="border: 1px solid black;">Modify [1, 2]</td>
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<td style="border: 1px solid black;">[-1, 0, 1]</td>
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<td style="border: 1px solid black;">(-1 &times; 4) + (0 &times; 2) + (1 &times; 8) = -4 + 0 + 8</td>
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<td style="border: 1px solid black;">4</td>
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</tr>
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</tbody>
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</table>
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<p>Thus, the maximum possible profit is 10, which is achieved by modifying the subarray <code>[0, 1]</code>​​​​​​​.</p>
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</div>
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<p><strong class="example">Example 2:</strong></p>
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<div class="example-block">
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<p><strong>Input:</strong> <span class="example-io">prices = [5,4,3], strategy = [1,1,0], k = 2</span></p>
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<p><strong>Output:</strong> <span class="example-io">9</span></p>
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<p><strong>Explanation:</strong></p>
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<div class="example-block">
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<table style="border: 1px solid black;">
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<thead>
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<tr>
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<th style="border: 1px solid black;">Modification</th>
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<th style="border: 1px solid black;">Strategy</th>
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<th style="border: 1px solid black;">Profit Calculation</th>
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<th style="border: 1px solid black;">Profit</th>
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</tr>
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</thead>
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<tbody>
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<tr>
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<td style="border: 1px solid black;">Original</td>
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<td style="border: 1px solid black;">[1, 1, 0]</td>
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<td style="border: 1px solid black;">(1 &times; 5) + (1 &times; 4) + (0 &times; 3) = 5 + 4 + 0</td>
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<td style="border: 1px solid black;">9</td>
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</tr>
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<tr>
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<td style="border: 1px solid black;">Modify [0, 1]</td>
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<td style="border: 1px solid black;">[0, 1, 0]</td>
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<td style="border: 1px solid black;">(0 &times; 5) + (1 &times; 4) + (0 &times; 3) = 0 + 4 + 0</td>
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<td style="border: 1px solid black;">4</td>
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</tr>
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<tr>
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<td style="border: 1px solid black;">Modify [1, 2]</td>
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<td style="border: 1px solid black;">[1, 0, 1]</td>
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<td style="border: 1px solid black;">(1 &times; 5) + (0 &times; 4) + (1 &times; 3) = 5 + 0 + 3</td>
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<td style="border: 1px solid black;">8</td>
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</tr>
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</tbody>
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</table>
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<p>Thus, the maximum possible profit is 9, which is achieved without any modification.</p>
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</div>
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</div>
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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>2 &lt;= prices.length == strategy.length &lt;= 10<sup>5</sup></code></li>
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<li><code>1 &lt;= prices[i] &lt;= 10<sup>5</sup></code></li>
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<li><code>-1 &lt;= strategy[i] &lt;= 1</code></li>
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<li><code>2 &lt;= k &lt;= prices.length</code></li>
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<li><code>k</code> is even</li>
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</ul>
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<!-- description:end -->
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## Solutions
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<!-- solution:start -->
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### Solution 1
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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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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->

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