<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, size):
        self.size = size
        self.tree = [0] * (size + 1)

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

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

class Solution:
    def goodTriplets(self, nums1: List[int], nums2: List[int]) -> int:
        n = len(nums1)
        
        # Map values in nums2 to their indices
        value_to_index = {value: index for index, value in enumerate(nums2)}
        
        # Convert nums1 to indices in nums2
        indices = [value_to_index[value] for value in nums1]
        
        # Initialize Fenwick Trees
        bit1 = FenwickTree(n)
        bit2 = FenwickTree(n)
        
        # Arrays to store the number of elements before and after each index
        left = [0] * n
        right = [0] * n
        
        # Calculate the number of elements before each index
        for i, index in enumerate(indices):
            left[i] = bit1.query(index + 1)
            bit1.update(index + 1, 1)
        
        # Calculate the number of elements after each index
        for i in range(n - 1, -1, -1):
            right[i] = bit2.query(n) - bit2.query(indices[i] + 1)
            bit2.update(indices[i] + 1, 1)
        
        # Calculate the number of good triplets
        good_triplets = 0
        for i in range(1, n - 1):
            good_triplets += left[i] * right[i]
        
        return good_triplets
1564