<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:
        n = len(nums)
        #1 32 4, i jk l
        right_3 = [0] * n # right of j
        right_3[-1] = [0] * (n + 1)
        for k in range(n - 2, 1, -1): 
            # right_3[k] = right_3[k+1].copy()
             right_3[k] = right_3[k + 1][:]
             for x in range(1, nums[k+1]):
                right_3[k][x] += 1
        
        ans = 0
        left_2 = [0] * (n + 1)# left of k
        for j in range(1,n - 1):
            for x in range(nums[j-1] + 1, n + 1):
                left_2[x] += 1
            for k in range(j + 1, n - 1):
                if nums[j] > nums[k]:
                    ans = ans + left_2[nums[k]] * right_3[k][nums[j]]#combination
        return ans
361900