<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>

class Solution:
    def countQuadruplets(self, nums: List[int]) -> int:
        # pre[i][j] = the number of index j <= x < i such that nums[x] < nums[i]
        # post[i][j] = the number of index j >= x > i such that nums[x] > nums[i]
        n = len(nums)
        pre = [[0 for _ in range(n)] for _ in range(n)]
        post = [[0 for _ in range(n)] for _ in range(n)]
        for i in range(n):
            pre[i][i] = 0
            for j in reversed(range(i)):
                pre[i][j] = pre[i][j + 1] + (1 if nums[j] < nums[i] else 0)
            
            post[i][i] = 0
            for j in range(i + 1, n):
                post[i][j] = post[i][j - 1] + (1 if nums[j] > nums[i] else 0)

        #print(pre)
        #print(post)

        ans = 0
        for j in range(n):
            for k in range(j + 1, n):
                if nums[j] <= nums[k]:
                    continue 
                # increment ans with (number of i < j such that nums[i] < nums[k]) * 
                # (number of l > k such that nums[l] > nums[j])
                ans += (pre[k][0] - pre[k][j]) * (post[j][n - 1] - post[j][k])
        
        return ans
        
653800