<p>You are given two <strong>0-indexed</strong> arrays <code>nums1</code> and <code>nums2</code> of length <code>n</code>, both of which are <strong>permutations</strong> of <code>[0, 1, ..., n - 1]</code>.</p>

<p>A <strong>good triplet</strong> is a set of <code>3</code> <strong>distinct</strong> values which are present in <strong>increasing order</strong> by position both in <code>nums1</code> and <code>nums2</code>. In other words, if we consider <code>pos1<sub>v</sub></code> as the index of the value <code>v</code> in <code>nums1</code> and <code>pos2<sub>v</sub></code> as the index of the value <code>v</code> in <code>nums2</code>, then a good triplet will be a set <code>(x, y, z)</code> where <code>0 &lt;= x, y, z &lt;= n - 1</code>, such that <code>pos1<sub>x</sub> &lt; pos1<sub>y</sub> &lt; pos1<sub>z</sub></code> and <code>pos2<sub>x</sub> &lt; pos2<sub>y</sub> &lt; pos2<sub>z</sub></code>.</p>

<p>Return <em>the <strong>total number</strong> of good triplets</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums1 = [2,0,1,3], nums2 = [0,1,2,3]
<strong>Output:</strong> 1
<strong>Explanation:</strong> 
There are 4 triplets (x,y,z) such that pos1<sub>x</sub> &lt; pos1<sub>y</sub> &lt; pos1<sub>z</sub>. They are (2,0,1), (2,0,3), (2,1,3), and (0,1,3). 
Out of those triplets, only the triplet (0,1,3) satisfies pos2<sub>x</sub> &lt; pos2<sub>y</sub> &lt; pos2<sub>z</sub>. Hence, there is only 1 good triplet.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums1 = [4,0,1,3,2], nums2 = [4,1,0,2,3]
<strong>Output:</strong> 4
<strong>Explanation:</strong> The 4 good triplets are (4,0,3), (4,0,2), (4,1,3), and (4,1,2).
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == nums1.length == nums2.length</code></li>
	<li><code>3 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= nums1[i], nums2[i] &lt;= n - 1</code></li>
	<li><code>nums1</code> and <code>nums2</code> are permutations of <code>[0, 1, ..., n - 1]</code>.</li>
</ul>

class FenwickTree:
    def __init__(self, n):
        self.tree = [0] * (n + 1)
        self.n = n

    def update(self, index, value):
        index += 1
        while index <= self.n:
            self.tree[index] += value
            index += index & -index

    def query(self, index):
        result = 0
        index += 1
        while index > 0:
            result += self.tree[index]
            index -= index & -index
        return result

    def range_query(self, left, right):
        return self.query(right) - self.query(left - 1)


class Solution:
    def goodTriplets(self, nums1: List[int], nums2: List[int]) -> int:
        
        n = len(nums1)
        
        # Step 1: Build position map for nums2
        pos2 = {num: i for i, num in enumerate(nums2)}
        
        # Step 2: Fenwick Trees to count the number of smaller and larger elements
        left_tree = FenwickTree(n)
        right_tree = FenwickTree(n)
        
        # To count how many triplets we can form
        triplet_count = 0
        
        # Step 3: Traverse nums1 and for each element, update counts
        # Since nums1 and nums2 are permutations, both arrays contain numbers 0 to n-1.
        # First, initialize the right Fenwick Tree with all elements in nums2
        for i in range(n):
            right_tree.update(pos2[nums1[i]], 1)
        
        # Step 4: Process each element in nums1
        for i in range(n):
            # Remove current element from the right Fenwick Tree
            right_tree.update(pos2[nums1[i]], -1)
            
            # Count elements smaller and larger
            smaller_count = left_tree.query(pos2[nums1[i]] - 1)
            larger_count = right_tree.query(n - 1) - right_tree.query(pos2[nums1[i]])
            
            # The number of good triplets involving nums1[i] is smaller_count * larger_count
            triplet_count += smaller_count * larger_count
            
            # Update the left Fenwick Tree with the current element
            left_tree.update(pos2[nums1[i]], 1)
        
        return triplet_count
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