<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, data, operation=min):
        self.n = len(data)
        self.tree = [0]*self.n + list(data)
        self.operation = operation
        for i in range(self.n - 1, 0, -1):
            self.tree[i] = self.operation(self.tree[i << 1], self.tree[i << 1 | 1])

    def update(self, index, value):
        pos = index + self.n
        self.tree[pos] = value
        while pos > 1:
            pos >>= 1
            self.tree[pos] = self.operation(self.tree[pos << 1], self.tree[pos << 1 | 1])

    def __getitem__(self, key):
        if isinstance(key, slice):
            start = key.start if key.start is not None else 0
            stop = key.stop if key.stop is not None else self.n
            return self.query(start, stop)
        elif isinstance(key, int) and key < self.n:
            return self.tree[key+self.n]
        else:
            raise TypeError(f"Value of {key} cannot be indexed.")

    def query(self, left, right):
        result = self.tree[left + self.n]
        left += self.n + 1
        right += self.n
        while left < right:
            if left & 1:
                result = self.operation(result, self.tree[left])
                left += 1
            if right & 1:
                right -= 1
                result = self.operation(result, self.tree[right])
            left >>= 1
            right >>= 1
        return result

class Solution:
    def goodTriplets(self, nums1: List[int], nums2: List[int]) -> int:

        m1 = [-1]*len(nums1)
        for i, val in enumerate(nums1):
            m1[val] = i
        m2 = [-1]*len(nums2)
        for i, val in enumerate(nums2):
            m2[val] = i
        nums = [(a, b) for a, b in zip(m1, m2)]
        nums = sorted(nums)

        smaller = []
        seg_tree = SegmentTree([0]*len(nums), operation=lambda x,y: x+y)
        for a, b in nums:
            smaller.append(seg_tree[0: b])
            seg_tree.update(b, 1)

        bigger = []
        seg_tree = SegmentTree([0]*len(nums), operation=lambda x,y: x+y)
        for a, b in reversed(nums):
            bigger.append(seg_tree[b:len(nums)])
            seg_tree.update(b, 1)
        bigger = bigger[::-1]

        ans = 0
        for s, b in zip(smaller, bigger):
            ans += s*b
        return ans
2315