<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 SegmentTree:
    def __init__(self, n):
        self.n = n
        self.seg = [0] * (4 * n + 10)
    
    def build(self, l, r, c=1):
        if l == r:
            self.seg[c] = 0
            return
        mid = (l + r) // 2
        self.build(l, mid, 2 * c)
        self.build(mid + 1, r, 2 * c + 1)
        self.seg[c] = self.seg[2 * c] + self.seg[2 * c + 1]

    def query(self, l, r, a, b, c=1):
        if l > b or r < a:
            return 0
        if l >= a and r <= b:
            return self.seg[c]
        mid = (l + r) // 2
        return self.query(l, mid, a, b, 2 * c) + self.query(mid + 1, r, a, b, 2 * c + 1)

    def update(self, l, r, ind, c=1):
        if ind < l or ind > r:
            return
        if l == r and l == ind:
            self.seg[c] = 1
            return
        mid = (l + r) // 2
        self.update(l, mid, ind, 2 * c)
        self.update(mid + 1, r, ind, 2 * c + 1)
        self.seg[c] = self.seg[2 * c] + self.seg[2 * c + 1]

class Solution:
    def goodTriplets(self, nums1, nums2):
        n = len(nums1)
        seg_tree = SegmentTree(n)
        seg_tree.build(1, n)
        
        ans = 0
        pos_in_nums2 = {val: idx + 1 for idx, val in enumerate(nums2)}
        
        seg_tree.update(1, n, pos_in_nums2[nums1[0]])

        for i in range(1, n):
            r = pos_in_nums2[nums1[i]]
            q = seg_tree.query(1, n, 1, r)
            d = i - q
            loc = (n - r) - d
            ans += loc * q
            
            seg_tree.update(1, n, r)
        
        return ans
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