<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 sortedcontainers import SortedList


class Solution:
    def goodTriplets(self, nums1: List[int], nums2: List[int]) -> int:
        sl1 = SortedList()
        sl2 = SortedList()
        prefix, suffix = [], []
        memo = {num: i for i, num in enumerate(nums2)}
        mapping = [memo[num] for num in nums1]
        n = len(nums1)
        for i in range(n):
            sl1.add(mapping[i])
            prefix.append(sl1.bisect_left(mapping[i]))
        for j in range(n - 1, -1, -1):
            sl2.add(mapping[j])
            suffix.append(len(sl2) - 1 - sl2.bisect_left(mapping[j]))
        # print(prefix, suffix)
        res = 0
        for i in range(n):
            res += prefix[i] * suffix[n - i - 1]
        return res


class BIT:
    def __init__(self, n):
        self.n = n
        self.tree = [0] * (n + 1)

    def update(self, index, val):
        i = index + 1
        while i < len(self.tree):
            self.tree[i] += val
            i += i & -i

    def query(self, i):
        s = 0
        while i:
            s += self.tree[i]
            i -= i & -i
        return s

    def sumRange(self, l, r):
        return self.query(r + 1) - self.query(l)



class Solution:
    def goodTriplets(self, nums1: List[int], nums2: List[int]) -> int:
        memo = {num: i for i, num in enumerate(nums2)}
        mapping = [memo[num] for num in nums1]
        n = len(nums1)
        prefix = BIT(n)
        suffix = BIT(n)
        prefix_cnt, suffix_cnt = [], []
        for i in range(n):
            prefix.update(mapping[i], 1)
            prefix_cnt.append(prefix.sumRange(0, mapping[i] - 1))
        for j in range(n - 1, -1, -1):
            suffix.update(mapping[j], 1)
            suffix_cnt.append(suffix.sumRange(mapping[j] + 1, n - 1))
        res = 0
        for i in range(n):
            res += prefix_cnt[i] * suffix_cnt[n - i - 1]
        return res

1743