Number of Ways to Reorder Array to Get Same BST
<p>Given an array <code>nums</code> that represents a permutation of integers from <code>1</code> to <code>n</code>. We are going to construct a binary search tree (BST) by inserting the elements of <code>nums</code> in order into an initially empty BST. Find the number of different ways to reorder <code>nums</code> so that the constructed BST is identical to that formed from the original array <code>nums</code>.</p>

<ul>
	<li>For example, given <code>nums = [2,1,3]</code>, we will have 2 as the root, 1 as a left child, and 3 as a right child. The array <code>[2,3,1]</code> also yields the same BST but <code>[3,2,1]</code> yields a different BST.</li>
</ul>

<p>Return <em>the number of ways to reorder</em> <code>nums</code> <em>such that the BST formed is identical to the original BST formed from</em> <code>nums</code>.</p>

<p>Since the answer may be very large, <strong>return it modulo </strong><code>10<sup>9</sup> + 7</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1500-1599/1569.Number%20of%20Ways%20to%20Reorder%20Array%20to%20Get%20Same%20BST/images/bb.png" style="width: 121px; height: 101px;" />
<pre>
<strong>Input:</strong> nums = [2,1,3]
<strong>Output:</strong> 1
<strong>Explanation:</strong> We can reorder nums to be [2,3,1] which will yield the same BST. There are no other ways to reorder nums which will yield the same BST.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1500-1599/1569.Number%20of%20Ways%20to%20Reorder%20Array%20to%20Get%20Same%20BST/images/ex1.png" style="width: 241px; height: 161px;" />
<pre>
<strong>Input:</strong> nums = [3,4,5,1,2]
<strong>Output:</strong> 5
<strong>Explanation:</strong> The following 5 arrays will yield the same BST: 
[3,1,2,4,5]
[3,1,4,2,5]
[3,1,4,5,2]
[3,4,1,2,5]
[3,4,1,5,2]
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1500-1599/1569.Number%20of%20Ways%20to%20Reorder%20Array%20to%20Get%20Same%20BST/images/ex4.png" style="width: 121px; height: 161px;" />
<pre>
<strong>Input:</strong> nums = [1,2,3]
<strong>Output:</strong> 0
<strong>Explanation:</strong> There are no other orderings of nums that will yield the same BST.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 1000</code></li>
	<li><code>1 &lt;= nums[i] &lt;= nums.length</code></li>
	<li>All integers in <code>nums</code> are <strong>distinct</strong>.</li>
</ul>
Tree; Union Find; Binary Search Tree; Memoization; Array; Math; Divide and Conquer; Dynamic Programming; Binary Tree; Combinatorics
class Solution:
    def numOfWays(self, nums: List[int]) -> int:
        def dfs(nums):
            if len(nums) < 2:
                return 1
            left = [x for x in nums if x < nums[0]]
            right = [x for x in nums if x > nums[0]]
            m, n = len(left), len(right)
            a, b = dfs(left), dfs(right)
            return (((c[m + n][m] * a) % mod) * b) % mod

        n = len(nums)
        mod = 10**9 + 7
        c = [[0] * n for _ in range(n)]
        c[0][0] = 1
        for i in range(1, n):
            c[i][0] = 1
            for j in range(1, i + 1):
                c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod
        return (dfs(nums) - 1 + mod) % mod
