<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:
        ans = 0
        for j, val_j in enumerate(nums):
            l = 0
            k_table = [0] * val_j
            for v in nums[-1:j:-1]:
                if v > val_j:
                    l += 1
                else:
                    k_table[v] = l
            k_table = list(accumulate(k_table[::-1]))[::-1]
            ans += sum(k_table[v] for v in nums[:j] if v < val_j)
        return ans
17400