<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 Ahnuf:
    def __init__(self, n):
        self.n = n
        self.bit = [0] * (n+1)

    def update(self, p):
        while p <= self.n: 
            self.bit[p] += 1
            p += (p & -p)

    def query(self, p):
        res = 0
        while p > 0:
            res += self.bit[p]
            p -= (p & -p)
        return res

class Solution:
    def clp(self, m1, m2):
        n = len(m1)
        i1, bit = {}, Ahnuf(n)
        
        for i in range(n):
            m = m1[i]
            i1[m] = i+1
        
        i2 = [i1[m] for m in m2]

        res = [0] * n
        for i in range(n):
            res[i] = bit.query(i2[i])
            bit.update(i2[i])

        return res
    
    def crp(self, m1, m2):
        n = len(m1)
        i1, bit = {}, Ahnuf(n)
        
        for i in range(n-1, -1, -1):
            m = m1[i]
            i1[m] = n-i
        
        i2 = [i1[m] for m in m2]

        res = [0] * n
        for i in range(n-1, -1, -1):
            res[i] = bit.query(i2[i])
            bit.update(i2[i])

        return res

    def goodTriplets(self, m1: List[int], m2: List[int]) -> int:
        res, n = 0, len(m1)
        lp = self.clp(m1, m2)
        rp = self.crp(m1, m2)
        for i in range(n):
            res += lp[i] * rp[i]
        return res
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