How do you solve Number of Ways to Reorder Array to Get Same BST on LeetCode?

Number of Ways to Reorder Array to Get Same BST (LeetCode #1569) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: The first element of the array is always the root of the BST.

What is the brute force solution for Number of Ways to Reorder Array to Get Same BST?

The brute force approach for Number of Ways to Reorder Array to Get Same BST is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible permutations of the array and checking which ones produce the same BST as the original. This is straightforward but inefficient due to the factorial growth of permutations. The optimal Optimal Solution approach improves this to O(n²).

What are the interview tips for Number of Ways to Reorder Array to Get Same BST?

Understand the properties of BSTs and how they relate to the problem. Practice combinatorial problems to get comfortable with calculating arrangements. Think about edge cases, such as arrays with only one or two elements.

What are common mistakes when solving Number of Ways to Reorder Array to Get Same BST?

Failing to correctly partition the array into left and right subtrees. Not using combinatorial math to count arrangements, leading to inefficient solutions.

#1569

Number of Ways to Reorder Array to Get Same BST

Hard

ArrayMathDivide and ConquerDynamic ProgrammingTreeUnion-FindBinary Search TreeMemoizationCombinatoricsBinary TreeDivide and ConquerDynamic ProgrammingCombinatorics

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n²)
Space	O(1)	O(n)

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Intuition

Time O(n²)Space O(n)

The optimal approach uses a divide-and-conquer strategy to count the number of valid BSTs that can be formed with the left and right subtrees of the root. It utilizes combinatorial mathematics to efficiently calculate the number of ways to arrange elements.

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Algorithm

4 steps

1Step 1: Identify the root (the first element of the array).
2Step 2: Partition the remaining elements into left and right subtrees based on their values relative to the root.
3Step 3: Recursively calculate the number of valid arrangements for the left and right subtrees.
4Step 4: Use combinatorial math (binomial coefficients) to calculate the number of ways to interleave the left and right subtree arrangements.

solution.py21 lines

1from math import comb
2
3MOD = 10**9 + 7
4
5def count_same_bst(nums):
6    def count_ways(start, end):
7        if start >= end:
8            return 1
9        root = nums[start]
10        left = [x for x in nums[start + 1:end + 1] if x < root]
11        right = [x for x in nums[start + 1:end + 1] if x > root]
12        left_count = count_ways(start + 1, start + len(left))
13        right_count = count_ways(start + len(left) + 1, end)
14        total_count = (left_count * right_count) % MOD
15        total_count = (total_count * comb(len(left) + len(right), len(left))) % MOD
16        return total_count
17
18    return count_ways(0, len(nums) - 1)
19
20# Example usage:
21print(count_same_bst([2, 1, 3]))  # Output: 1

ℹ

Complexity note: The time complexity is O(n²) due to the recursive calls for counting left and right subtrees, while the space complexity is O(n) due to the recursion stack.

1The first element of the array is always the root of the BST.
2The arrangement of left and right subtrees can be calculated using combinatorial mathematics.

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