LeetCode in Kotlin

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

Hard

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

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

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

Since the answer may be very large, return it modulo 109 + 7.

Example 1:

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:

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:

Input: nums = [1,2,3]

Output: 0

Explanation: There are no other orderings of nums that will yield the same BST.

Constraints:

• 1 <= nums.length <= 1000
• 1 <= nums[i] <= nums.length
• All integers in nums are distinct.

Solution

class Solution {
    fun numOfWays(nums: IntArray): Int {
        val mod: Long = 1000000007
        val fact = LongArray(1001)
        fact[0] = 1
        for (i in 1..1000) {
            fact[i] = fact[i - 1] * i % mod
        }
        val root = TreeNode(nums[0])
        for (i in 1 until nums.size) {
            addInTree(nums[i], root)
        }
        return ((calcPerms(root, fact).perm - 1) % mod).toInt()
    }

    class Inverse(var x: Long, var y: Long)
    class TreeInfo(var numOfNodes: Long, var perm: Long)
    class TreeNode(var `val`: Int) {
        var left: TreeNode? = null
        var right: TreeNode? = null
    }

    private fun calcPerms(root: TreeNode?, fact: LongArray): TreeInfo {
        val left: TreeInfo
        val right: TreeInfo
        left = if (root!!.left != null) {
            calcPerms(
                root.left, fact
            )
        } else {
            TreeInfo(0, 1)
        }
        right = if (root.right != null) {
            calcPerms(
                root.right, fact
            )
        } else {
            TreeInfo(0, 1)
        }
        val mod: Long = 1000000007
        val totNodes = left.numOfNodes + right.numOfNodes + 1
        val modDiv = getModDivision(
            fact[totNodes.toInt() - 1],
            fact[left.numOfNodes.toInt()],
            fact[right.numOfNodes.toInt()],
            mod
        )
        val perms = if (totNodes == 1L) 1 else left.perm * right.perm % mod * modDiv % mod
        left.numOfNodes = totNodes
        left.perm = perms
        return left
    }

    private fun getModDivision(a: Long, b1: Long, b2: Long, m: Long): Long {
        val b = b1 * b2
        val inv = getInverse(b, m)
        return inv * a % m
    }

    private fun getInverse(b: Long, m: Long): Long {
        val inv = getInverseExtended(b, m)
        return (inv.x % m + m) % m
    }

    private fun getInverseExtended(a: Long, b: Long): Inverse {
        if (a == 0L) {
            return Inverse(0, 1)
        }
        val inv = getInverseExtended(b % a, a)
        val x1 = inv.y - b / a * inv.x
        val y1 = inv.x
        inv.x = x1
        inv.y = y1
        return inv
    }

    private fun addInTree(x: Int, root: TreeNode?) {
        if (root!!.`val` > x) {
            if (root.left != null) {
                addInTree(x, root.left)
            } else {
                root.left = TreeNode(x)
            }
        }
        if (root.`val` < x) {
            if (root.right != null) {
                addInTree(x, root.right)
            } else {
                root.right = TreeNode(x)
            }
        }
    }
}

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