Count Increasing Quadruplets
<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>
Binary Indexed Tree; Array; Dynamic Programming; Enumeration; Prefix Sum
func countQuadruplets(nums []int) int64 {
	n := len(nums)
	f := make([][]int, n)
	g := make([][]int, n)
	for i := range f {
		f[i] = make([]int, n)
		g[i] = make([]int, n)
	}
	for j := 1; j < n-2; j++ {
		cnt := 0
		for l := j + 1; l < n; l++ {
			if nums[l] > nums[j] {
				cnt++
			}
		}
		for k := j + 1; k < n-1; k++ {
			if nums[j] > nums[k] {
				f[j][k] = cnt
			} else {
				cnt--
			}
		}
	}
	ans := 0
	for k := 2; k < n-1; k++ {
		cnt := 0
		for i := 0; i < k; i++ {
			if nums[i] < nums[k] {
				cnt++
			}
		}
		for j := k - 1; j > 0; j-- {
			if nums[j] > nums[k] {
				g[j][k] = cnt
				ans += f[j][k] * g[j][k]
			} else {
				cnt--
			}
		}
	}
	return int64(ans)
}