# LC 1569. Number of Ways to Reorder Array to Get Same BST

### [Problem link](https://leetcode.com/problems/number-of-ways-to-reorder-array-to-get-same-bst/)

###### tags: `leedcode` `python` `hard` `Math`

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>.

- 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.

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

Since the answer may be very large, **return it modulo ** <code>10⁹ + 7</code>.

**Example 1:**
<img alt="" src="https://assets.leetcode.com/uploads/2020/08/12/bb.png" style="width: 121px; height: 101px;" />

Input: nums = [2,1,3]
Output: 1
Explanation: 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.

**Example 2:**
<img alt="" src="https://assets.leetcode.com/uploads/2020/08/12/ex1.png" style="width: 241px; height: 161px;" />

Input: nums = [3,4,5,1,2]
Output: 5
Explanation: 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]

**Example 3:**
<img alt="" src="https://assets.leetcode.com/uploads/2020/08/12/ex4.png" style="width: 121px; height: 161px;" />

Input: nums = [1,2,3]
Output: 0
Explanation: There are no other orderings of nums that will yield the same BST.

**Constraints:**

- <code>1 <= nums.length <= 1000</code>
- <code>1 <= nums[i] <= nums.length</code>
- All integers in <code>nums</code> are **distinct** .

## Solution 1 - Math

class Solution:
    def numOfWays(self, nums: List[int]) -> int:
        mod = 10 ** 9 + 7
        n = len(nums)

        combMemo = [[0 for _ in range(n)] for _ in range(n)]
        def calComb(m, n):
            '''
            calculate C(m, n)
            '''
            if combMemo[m][n] != 0:
                return combMemo[m][n]
            if n > m - n:
                return calComb(m, m - n)
            if n == 0: return 1
            if n == 1: return m

            combMemo[m][n] = (calComb(m - 1, n) + calComb(m - 1, n - 1)) % mod

            return combMemo[m][n]

        def dfs(nums):
            m = len(nums)
            
            if m < 3:
                return 1

            left_nodes, right_nodes = [], []
            for num in nums[1:]:
                if num < nums[0]:
                    left_nodes.append(num)
                else:
                    right_nodes.append(num)

            left_val = dfs(left_nodes)
            right_val = dfs(right_nodes)

            return left_val * right_val * calComb(m - 1, len(left_nodes))

        return (dfs(nums) - 1) % mod

>### Complexity
>n = nums.length
>| | Time Complexity | Space Complexity |
>| ----------- | --------------- | ---------------- |
>| Solution 1 | O($n^2$) | O($n^2$) |

## Note
計算combination時也可以用math.comb(), time using會差非常多.

[reference(video)](https://www.youtube.com/watch?v=FxGWaG9danM)