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leetcode-problemset/leetcode-cn/problem (English)/统计上升四元组(English) [count-increasing-quadruplets].html
2023-02-02 20:54:33 +08:00

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<p>Given a <strong>0-indexed</strong> integer array <code>nums</code> of size <code>n</code> containing all numbers from <code>1</code> to <code>n</code>, return <em>the number of increasing quadruplets</em>.</p>
<p>A quadruplet <code>(i, j, k, l)</code> is increasing if:</p>
<ul>
<li><code>0 &lt;= i &lt; j &lt; k &lt; l &lt; n</code>, and</li>
<li><code>nums[i] &lt; nums[k] &lt; nums[j] &lt; nums[l]</code>.</li>
</ul>
<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre>
<strong>Input:</strong> nums = [1,3,2,4,5]
<strong>Output:</strong> 2
<strong>Explanation:</strong>
- When i = 0, j = 1, k = 2, and l = 3, nums[i] &lt; nums[k] &lt; nums[j] &lt; nums[l].
- When i = 0, j = 1, k = 2, and l = 4, nums[i] &lt; nums[k] &lt; nums[j] &lt; nums[l].
There are no other quadruplets, so we return 2.
</pre>
<p><strong class="example">Example 2:</strong></p>
<pre>
<strong>Input:</strong> nums = [1,2,3,4]
<strong>Output:</strong> 0
<strong>Explanation:</strong> There exists only one quadruplet with i = 0, j = 1, k = 2, l = 3, but since nums[j] &lt; nums[k], we return 0.
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>
<ul>
<li><code>4 &lt;= nums.length &lt;= 4000</code></li>
<li><code>1 &lt;= nums[i] &lt;= nums.length</code></li>
<li>All the integers of <code>nums</code> are <strong>unique</strong>. <code>nums</code> is a permutation.</li>
</ul>