<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>

from typing import List

class Solution:
    def goodTriplets(self, nums1: List[int], nums2: List[int]) -> int:
        n = len(nums1)
        
        # To store the position of each element of nums2 in nums1
        pos_in_nums2 = [0] * n
        for i in range(n):
            pos_in_nums2[nums2[i]] = i

        # Fenwick Tree to count increasing pairs (x, y) in nums1 corresponding to their positions in nums2.
        BITree1 = [0] * (n + 1)
        BITree2 = [0] * (n + 1)

        def update(BITree, idx, val):
            idx += 1  # BITree is 1-indexed
            while idx <= n:
                BITree[idx] += val
                idx += idx & (-idx)

        def query(BITree, idx):
            sum = 0
            idx += 1  # BITree is 1-indexed
            while idx > 0:
                sum += BITree[idx]
                idx -= idx & (-idx)
            return sum

        total_good_triplets = 0

        # Iterate nums1 and count triplets
        for i in range(n):
            pos = pos_in_nums2[nums1[i]]

            # Count of values before `nums1[i]` in terms of their positions in nums2
            count_before_in_nums2 = query(BITree1, pos - 1)  # sum up to the position before current

            # Update total_triplets count by sum of second BITree up to position
            total_good_triplets += query(BITree2, pos - 1)

            # Update the first BITree for position frequency
            update(BITree1, pos, 1)
            
            # Update the second BITree using count of previous pairs forming triplets with current `nums1[i]`
            update(BITree2, pos, count_before_in_nums2)

        return total_good_triplets
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