leetcode-problemset/leetcode-cn/originData/count-increasing-quadruplets.json

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            "title": "Count Increasing Quadruplets",
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            "content": "<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>\n\n<p>A quadruplet <code>(i, j, k, l)</code> is increasing if:</p>\n\n<ul>\n\t<li><code>0 &lt;= i &lt; j &lt; k &lt; l &lt; n</code>, and</li>\n\t<li><code>nums[i] &lt; nums[k] &lt; nums[j] &lt; nums[l]</code>.</li>\n</ul>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,3,2,4,5]\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> \n- When i = 0, j = 1, k = 2, and l = 3, nums[i] &lt; nums[k] &lt; nums[j] &lt; nums[l].\n- When i = 0, j = 1, k = 2, and l = 4, nums[i] &lt; nums[k] &lt; nums[j] &lt; nums[l]. \nThere are no other quadruplets, so we return 2.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,2,3,4]\n<strong>Output:</strong> 0\n<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.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>4 &lt;= nums.length &lt;= 4000</code></li>\n\t<li><code>1 &lt;= nums[i] &lt;= nums.length</code></li>\n\t<li>All the integers of <code>nums</code> are <strong>unique</strong>. <code>nums</code> is a permutation.</li>\n</ul>\n",
            "translatedTitle": "统计上升四元组",
            "translatedContent": "<p>给你一个长度为 <code>n</code>&nbsp;下标从 <strong>0</strong>&nbsp;开始的整数数组&nbsp;<code>nums</code>&nbsp;，它包含&nbsp;<code>1</code>&nbsp;到&nbsp;<code>n</code>&nbsp;的所有数字，请你返回上升四元组的数目。</p>\n\n<p>如果一个四元组&nbsp;<code>(i, j, k, l)</code>&nbsp;满足以下条件，我们称它是上升的：</p>\n\n<ul>\n\t<li><code>0 &lt;= i &lt; j &lt; k &lt; l &lt; n</code>&nbsp;且</li>\n\t<li><code>nums[i] &lt; nums[k] &lt; nums[j] &lt; nums[l]</code>&nbsp;。</li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre><b>输入：</b>nums = [1,3,2,4,5]\n<b>输出：</b>2\n<b>解释：</b>\n- 当 i = 0 ，j = 1 ，k = 2 且 l = 3 时，有 nums[i] &lt; nums[k] &lt; nums[j] &lt; nums[l] 。\n- 当 i = 0 ，j = 1 ，k = 2 且 l = 4 时，有 nums[i] &lt; nums[k] &lt; nums[j] &lt; nums[l] 。\n没有其他的四元组，所以我们返回 2 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre><b>输入：</b>nums = [1,2,3,4]\n<b>输出：</b>0\n<b>解释：</b>只存在一个四元组 i = 0 ，j = 1 ，k = 2 ，l = 3 ，但是 nums[j] &lt; nums[k] ，所以我们返回 0 。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>4 &lt;= nums.length &lt;= 4000</code></li>\n\t<li><code>1 &lt;= nums[i] &lt;= nums.length</code></li>\n\t<li><code>nums</code>&nbsp;中所有数字 <strong>互不相同</strong>&nbsp;，<code>nums</code>&nbsp;是一个排列。</li>\n</ul>\n",
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                "We can pre-compute all possible (i, j) and (k, l) and store them in 2 matrices.",
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