#1569

Hard

auto_awesomeBinary-tree traversal and state tracking

LeetCode Problem Workspace

Number of Ways to Reorder Array to Get Same BST

Determine the number of ways to reorder an array to get the same binary search tree (BST) from its insertion order.

array math divide and conquer dynamic programming tree

Problem Statement

You are given an array nums that represents a permutation of integers from 1 to n. By inserting these integers into an initially empty binary search tree (BST), you will construct a BST. Your task is to find how many different ways you can reorder the elements of nums such that the constructed BST is identical to the BST formed by the original order of nums.

The answer may be large, so return it modulo 10^9 + 7. If there are no reorderings that form the same BST, return 0. The problem involves binary-tree traversal, dynamic programming, and combinatorics to calculate valid reorderings while efficiently managing subproblems.

Examples

Example 1

Input: nums = [2,1,3]

Output: 1

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

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

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 Approach

Recursive Binary Tree Construction

Recursively construct the BST from the given array. For each subtree, track the possible positions for elements based on their order of insertion. This helps to break the problem down into smaller subproblems.

Combinatorial Counting

For each subtree, use combinatorics to count how many different ways elements can be ordered while maintaining the BST properties. Use binomial coefficients to calculate valid reorderings of elements within each subtree.

Memoization and Dynamic Programming

Store intermediate results using memoization to avoid redundant calculations, improving time complexity. This allows efficient computation of valid reorderings for each possible subtree structure.

Complexity Analysis

Metric	Value
Time	O(m^2)
Space	O(m^2)

The time complexity is O(m^2) due to recursive function calls and combinatorial counting with binomial coefficients. The space complexity is O(m^2) due to memoization and storing intermediate results for subproblems, where m is the length of the input array.

What Interviewers Usually Probe

Candidate shows strong understanding of divide and conquer techniques in solving tree-based problems.
Candidate demonstrates the ability to optimize recursive solutions using dynamic programming and memoization.
Candidate is able to efficiently apply combinatorics to count valid reorderings within subarrays of the BST.

Common Pitfalls or Variants

Common pitfalls

Failing to account for all possible reorderings within each subtree.
Not properly using memoization, leading to redundant recalculations.
Incorrectly calculating binomial coefficients or failing to apply them in subtree reorderings.

Follow-up variants

Handling larger tree structures with increased complexity in counting reorderings.
Introducing additional constraints, such as restricting specific numbers to particular positions.
Optimizing for space complexity, especially when dealing with very large arrays.

FAQ

What is the primary approach to solving the Number of Ways to Reorder Array to Get Same BST problem?

The primary approach is to recursively divide the array into subarrays while maintaining the binary search tree properties, then apply combinatorics to count valid reorderings.

How does the divide and conquer pattern apply to this problem?

Divide and conquer is used to break the problem into smaller subproblems by recursively constructing the BST and counting valid reorderings within each subtree.

What role does dynamic programming play in solving this problem?

Dynamic programming is used to store intermediate results and avoid redundant calculations, speeding up the solution by reusing previously computed values.

How do binomial coefficients help in counting valid reorderings of the array?

Binomial coefficients are used to calculate how many ways elements can be reordered within a subtree while preserving the binary search tree structure.

What happens if there are no reorderings that can form the same BST?

If no reorderings are possible, the function should return 0, indicating that no valid permutations exist that maintain the same BST structure.

terminal

Solution

Solution 1: Combination Counting + Recursion

We design a function $dfs(nums)$, which is used to calculate the number of solutions of the binary search tree with $nums$ as nodes. Then the answer is $dfs(nums)-1$, because $dfs(nums)$ calculates the number of solutions of the binary search tree with $nums$ as nodes, while the problem requires the number of solutions of the binary search tree with $nums$ as nodes after reordering, so the answer needs to be subtracted by one.

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

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

Continue Practicing

#1467 probability of a two boxes having the same number of distinct balls #1728 cat and mouse ii #1382 balance a binary search tree