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Commit bc4695b

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yanglbmeidoocs
andauthored
feat: add new lc problems: No.2873~2876 (#1733)
* No.2873.Maximum Value of an Ordered Triplet I * No.2874.Maximum Value of an Ordered Triplet II * No.2875.Minimum Size Subarray in Infinite Array * No.2876.Count Visited Nodes in a Directed Graph --------- Co-authored-by: Doocs Bot <doocs-bot@outlook.com>
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# [2873. 有序三元组中的最大值 I](https://leetcode.cn/problems/maximum-value-of-an-ordered-triplet-i)
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[English Version](/solution/2800-2899/2873.Maximum%20Value%20of%20an%20Ordered%20Triplet%20I/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>给你一个下标从 <strong>0</strong> 开始的整数数组 <code>nums</code> 。</p>
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<p>请你从所有满足&nbsp;<code>i &lt; j &lt; k</code> 的下标三元组 <code>(i, j, k)</code> 中,找出并返回下标三元组的最大值。如果所有满足条件的三元组的值都是负数,则返回 <code>0</code> 。</p>
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<p><strong>下标三元组</strong> <code>(i, j, k)</code> 的值等于 <code>(nums[i] - nums[j]) * nums[k]</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>nums = [12,6,1,2,7]
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<strong>输出:</strong>77
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<strong>解释:</strong>下标三元组 (0, 2, 4) 的值是 (nums[0] - nums[2]) * nums[4] = 77 。
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可以证明不存在值大于 77 的有序下标三元组。
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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>nums = [1,10,3,4,19]
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<strong>输出:</strong>133
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<strong>解释:</strong>下标三元组 (1, 2, 4) 的值是 (nums[1] - nums[2]) * nums[4] = 133 。
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可以证明不存在值大于 133 的有序下标三元组。
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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>nums = [1,2,3]
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<strong>输出:</strong>0
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<strong>解释:</strong>唯一的下标三元组 (0, 1, 2) 的值是一个负数,(nums[0] - nums[1]) * nums[2] = -3 。因此,答案是 0 。
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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>3 &lt;= nums.length &lt;= 100</code></li>
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<li><code>1 &lt;= nums[i] &lt;= 10<sup>6</sup></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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# [2873. Maximum Value of an Ordered Triplet I](https://leetcode.com/problems/maximum-value-of-an-ordered-triplet-i)
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[中文文档](/solution/2800-2899/2873.Maximum%20Value%20of%20an%20Ordered%20Triplet%20I/README.md)
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## Description
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<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code>.</p>
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<p>Return <em><strong>the maximum value over all triplets of indices</strong></em> <code>(i, j, k)</code> <em>such that</em> <code>i &lt; j &lt; k</code>. If all such triplets have a negative value, return <code>0</code>.</p>
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<p>The <strong>value of a triplet of indices</strong> <code>(i, j, k)</code> is equal to <code>(nums[i] - nums[j]) * nums[k]</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> nums = [12,6,1,2,7]
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<strong>Output:</strong> 77
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<strong>Explanation:</strong> The value of the triplet (0, 2, 4) is (nums[0] - nums[2]) * nums[4] = 77.
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It can be shown that there are no ordered triplets of indices with a value greater than 77.
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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> nums = [1,10,3,4,19]
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<strong>Output:</strong> 133
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<strong>Explanation:</strong> The value of the triplet (1, 2, 4) is (nums[1] - nums[2]) * nums[4] = 133.
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It can be shown that there are no ordered triplets of indices with a value greater than 133.
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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> nums = [1,2,3]
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<strong>Output:</strong> 0
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<strong>Explanation:</strong> The only ordered triplet of indices (0, 1, 2) has a negative value of (nums[0] - nums[1]) * nums[2] = -3. Hence, the answer would be 0.
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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>3 &lt;= nums.length &lt;= 100</code></li>
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<li><code>1 &lt;= nums[i] &lt;= 10<sup>6</sup></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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# [2874. 有序三元组中的最大值 II](https://leetcode.cn/problems/maximum-value-of-an-ordered-triplet-ii)
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[English Version](/solution/2800-2899/2874.Maximum%20Value%20of%20an%20Ordered%20Triplet%20II/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>给你一个下标从 <strong>0</strong> 开始的整数数组 <code>nums</code> 。</p>
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<p>请你从所有满足&nbsp;<code>i &lt; j &lt; k</code> 的下标三元组 <code>(i, j, k)</code> 中,找出并返回下标三元组的最大值。如果所有满足条件的三元组的值都是负数,则返回 <code>0</code> 。</p>
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<p><strong>下标三元组</strong> <code>(i, j, k)</code> 的值等于 <code>(nums[i] - nums[j]) * nums[k]</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>nums = [12,6,1,2,7]
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<strong>输出:</strong>77
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<strong>解释:</strong>下标三元组 (0, 2, 4) 的值是 (nums[0] - nums[2]) * nums[4] = 77 。
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可以证明不存在值大于 77 的有序下标三元组。
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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>nums = [1,10,3,4,19]
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<strong>输出:</strong>133
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<strong>解释:</strong>下标三元组 (1, 2, 4) 的值是 (nums[1] - nums[2]) * nums[4] = 133 。
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可以证明不存在值大于 133 的有序下标三元组。
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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>nums = [1,2,3]
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<strong>输出:</strong>0
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<strong>解释:</strong>唯一的下标三元组 (0, 1, 2) 的值是一个负数,(nums[0] - nums[1]) * nums[2] = -3 。因此,答案是 0 。
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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>3 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
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<li><code>1 &lt;= nums[i] &lt;= 10<sup>6</sup></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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# [2874. Maximum Value of an Ordered Triplet II](https://leetcode.com/problems/maximum-value-of-an-ordered-triplet-ii)
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[中文文档](/solution/2800-2899/2874.Maximum%20Value%20of%20an%20Ordered%20Triplet%20II/README.md)
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## Description
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<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code>.</p>
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<p>Return <em><strong>the maximum value over all triplets of indices</strong></em> <code>(i, j, k)</code> <em>such that</em> <code>i &lt; j &lt; k</code><em>. </em>If all such triplets have a negative value, return <code>0</code>.</p>
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<p>The <strong>value of a triplet of indices</strong> <code>(i, j, k)</code> is equal to <code>(nums[i] - nums[j]) * nums[k]</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> nums = [12,6,1,2,7]
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<strong>Output:</strong> 77
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<strong>Explanation:</strong> The value of the triplet (0, 2, 4) is (nums[0] - nums[2]) * nums[4] = 77.
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It can be shown that there are no ordered triplets of indices with a value greater than 77.
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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> nums = [1,10,3,4,19]
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<strong>Output:</strong> 133
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<strong>Explanation:</strong> The value of the triplet (1, 2, 4) is (nums[1] - nums[2]) * nums[4] = 133.
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It can be shown that there are no ordered triplets of indices with a value greater than 133.
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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> nums = [1,2,3]
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<strong>Output:</strong> 0
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<strong>Explanation:</strong> The only ordered triplet of indices (0, 1, 2) has a negative value of (nums[0] - nums[1]) * nums[2] = -3. Hence, the answer would be 0.
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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>3 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
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<li><code>1 &lt;= nums[i] &lt;= 10<sup>6</sup></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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