What is the time complexity of Number of Ways to Reorder Array to Get Same BST?

The optimal solution has a time complexity of O(n^2).

What are the key insights for solving Number of Ways to Reorder Array to Get Same BST?

Insight 1: The root of the BST will always be the first element of the array. The remaining elements can be divided into two groups: those smaller than the root (left subtree) and those larger than the root (right subtree). Insight 2: The order of elements within the left and right subtrees doesn't affect the BST structure, but their relative order with respect to each other (left elements before right elements) does.

Which companies ask LeetCode 1569. Number of Ways to Reorder Array to Get Same BST in interviews?

This problem is frequently asked at DE Shaw.

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Number of Ways to Reorder Array to Get Same BST - LeetCode 1569 Solution

Q: What is the space complexity of Number of Ways to Reorder Array to Get Same BST?

The space complexity is O(n^2).

Q: What data structures are used to solve Number of Ways to Reorder Array to Get Same BST?

This problem uses Array, Math, Divide and Conquer, Dynamic Programming, Tree, Union Find, Binary Search Tree, Memoization, Combinatorics, Binary Tree. Understanding these concepts is essential for solving this Hard problem efficiently.

Number of Ways to Reorder Array to Get Same BST - Complete Solution Guide

Number of Ways to Reorder Array to Get Same BST is LeetCode problem 1569, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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]

Detailed Explanation

The problem asks us to find the number of ways to reorder an array `nums` such that the binary search tree (BST) constructed by inserting the elements of the reordered array into an initially empty BST is identical to the BST constructed from the original array `nums`. The array `nums` contains a permutation of integers from 1 to n. We need to return the number of such reorderings modulo 10^9 + 7.

Solution Approach

The solution uses a recursive approach combined with combinatorics to calculate the number of valid reorderings. It first builds a helper function `dfs` that takes an array as input. This function recursively calculates the number of ways to reorder the given array such that it results in the same BST. Inside the `dfs` function, it divides the array into left and right subtrees based on the root node (the first element). Then, it uses combinations to calculate the number of ways to interleave the left and right subtrees. Finally, it recursively calls `dfs` on the left and right subtrees and multiplies the results together, modulo 10^9 + 7. A combination table is precomputed using dynamic programming to calculate combination values efficiently. The initial call to `dfs` is made with the original `nums` array and we subtract 1 from the final result to exclude the original array itself.

Step-by-Step Algorithm

Step 1: Precompute the combinations table C[i][j] using dynamic programming. This table will store the values of 'i choose j' modulo 10^9 + 7.
Step 2: Define a recursive function `dfs(arr)` that takes an array `arr` as input.
Step 3: Base Case: If the length of `arr` is less than or equal to 2, return 1 (as there is only one way to reorder).
Step 4: Divide the array `arr` into two subarrays: `left_nodes` containing elements smaller than the root (arr[0]) and `right_nodes` containing elements larger than the root.
Step 5: Calculate the number of ways to interleave the left and right subtrees using combinations: `C[len(arr) - 1][len(left_nodes)]`.
Step 6: Recursively calculate the number of ways to reorder the left and right subtrees using `dfs(left_nodes)` and `dfs(right_nodes)`.
Step 7: Multiply the number of ways to interleave, the number of ways to reorder the left subtree, and the number of ways to reorder the right subtree, all modulo 10^9 + 7. This gives the number of ways to reorder `arr`.
Step 8: Return the result from the recursive function.
Step 9: In the main function, call the `dfs` function with the input array `nums`.
Step 10: Subtract 1 from the result (to exclude the original array) and take the modulo 10^9 + 7. This handles the edge case where we don't want to count the original array.

Key Insights

Insight 1: The root of the BST will always be the first element of the array. The remaining elements can be divided into two groups: those smaller than the root (left subtree) and those larger than the root (right subtree).
Insight 2: The order of elements within the left and right subtrees doesn't affect the BST structure, but their relative order with respect to each other (left elements before right elements) does.
Insight 3: We can use combinations to count the number of ways to interleave the left and right subtrees. The formula for combinations (nCr) is crucial for calculating the possible arrangements.
Insight 4: Recursion and Divide and Conquer are suitable techniques to solve this problem because the subtrees can be solved independently and the results can be combined.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n^2)

Topics

This problem involves: Array, Math, Divide and Conquer, Dynamic Programming, Tree, Union Find, Binary Search Tree, Memoization, Combinatorics, Binary Tree.

Companies

Asked at: DE Shaw.

Number of Ways to Reorder Array to Get Same BST | LeetCode 1569

Hard

Array Math Divide and Conquer Dynamic Programming Tree Union Find Binary Search Tree Memoization Combinatorics Binary Tree

Time: O(n^2)Space: O(n^2)

Solution Code

Available in:

class Solution {
    private static final int MOD = 1000000007;

    public int numOfWays(int[] nums) {
        int n = nums.length;

        long[][] C = new long[n + 1][n + 1];
        for (int i = 0; i <= n; i++) {
            C[i][0] = 1;
            for (int j = 1; j <= i; j++) {
                C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % MOD;
            }
        }

        return (int)((dfs(nums, C) - 1 + MOD) % MOD);
    }

    private long dfs(int[] arr, long[][] C) {
        int m = arr.length;
        if (m <= 2) {
            return 1;
        }

        int root = arr[0];
        int[] left_nodes = new int[0];
        int[] right_nodes = new int[0];
        
        int leftCount = 0;
        for(int x : arr){
            if(x < root){
                leftCount++;
            }
        }

        int rightCount = 0;
         for(int x : arr){
            if(x > root){
                rightCount++;
            }
        }

        left_nodes = new int[leftCount];
        right_nodes = new int[rightCount];

        int leftIndex = 0;
        for (int x : arr) {
            if (x < root) {
                left_nodes[leftIndex++] = x;
            }
        }
        
        int rightIndex = 0;
         for (int x : arr) {
            if (x > root) {
                right_nodes[rightIndex++] = x;
            }
        }

        int len_left = left_nodes.length;

        long ways_to_interleave = C[m - 1][len_left];

        long ways_left = dfs(left_nodes, C);
        long ways_right = dfs(right_nodes, C);

        long result = (ways_to_interleave * ways_left) % MOD;
        result = (result * ways_right) % MOD;
        return result;
    }
}

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Problem Info

Description

Problem 1569: Number of Ways to Reorder Array to Get Same BST - Detailed explanation and solution approaches.

Topics

Array Math Divide and Conquer Dynamic Programming Tree Union Find Binary Search Tree Memoization Combinatorics Binary Tree

Companies

DE Shaw

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1569

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

10 steps

1
Step 1: Precompute the combinations table C[i][j] using dynamic programming. This table will store the values of 'i choose j' modulo 10^9 + 7.
2
Step 2: Define a recursive function `dfs(arr)` that takes an array `arr` as input.
3
Step 3: Base Case: If the length of `arr` is less than or equal to 2, return 1 (as there is only one way to reorder).
4
Step 4: Divide the array `arr` into two subarrays: `left_nodes` containing elements smaller than the root (arr[0]) and `right_nodes` containing elements larger than the root.
5
Step 5: Calculate the number of ways to interleave the left and right subtrees using combinations: `C[len(arr) - 1][len(left_nodes)]`.
6
Step 6: Recursively calculate the number of ways to reorder the left and right subtrees using `dfs(left_nodes)` and `dfs(right_nodes)`.
7
Step 7: Multiply the number of ways to interleave, the number of ways to reorder the left subtree, and the number of ways to reorder the right subtree, all modulo 10^9 + 7. This gives the number of ways to reorder `arr`.
8
Step 8: Return the result from the recursive function.
9
Step 9: In the main function, call the `dfs` function with the input array `nums`.
10
Step 10: Subtract 1 from the result (to exclude the original array) and take the modulo 10^9 + 7. This handles the edge case where we don't want to count the original array.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(n^2)

📚 Quadratic space - Uses lots of memory!

📝

Why?

The time complexity is O(n^2) because of two main factors: 1. Building the combinations table C, which takes O(n^2) time. 2. The `dfs` function can be called at most n times (depth of the BST). Inside the `dfs` function, creating the left and right subarrays takes O(n) time. Also, the combination calculation is considered O(1) since we have precomputed it. The space complexity is O(n^2) because the combinations table C takes O(n^2) space. Additionally, the recursive calls can take up to O(n) space on the call stack in the worst-case scenario (skewed BST).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Iterative Approach using Dynamic Programming

Instead of recursion, we could potentially use an iterative dynamic programming approach. However, this is complex to implement and would likely have a higher space complexity due to the need to store intermediate results for subtrees.

Time: Potentially O(n^3) or higherSpace: Potentially O(n^3) or higher

Trade-offs: The recursive approach is generally easier to understand and implement. The iterative approach is probably not beneficial for this problem given the constraints.

✨

Key Insights

1Insight 1: The root of the BST will always be the first element of the array. The remaining elements can be divided into two groups: those smaller than the root (left subtree) and those larger than the root (right subtree).
2Insight 2: The order of elements within the left and right subtrees doesn't affect the BST structure, but their relative order with respect to each other (left elements before right elements) does.
3Insight 3: We can use combinations to count the number of ways to interleave the left and right subtrees. The formula for combinations (nCr) is crucial for calculating the possible arrangements.
4Insight 4: Recursion and Divide and Conquer are suitable techniques to solve this problem because the subtrees can be solved independently and the results can be combined.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: `nums` contains only one element. The code handles this case correctly because the base case `m <= 2` will return 1, which represents 1 way to reorder the array.
Edge case 2: The array is already sorted (ascending or descending). In this case, there is only one possible BST structure and the result should be 0 (since the original array should be excluded). Subtracting 1 from the total ways ensures this is handled correctly.
Edge case 3: Large input array size close to the limit (n=1000). The precomputed combinations table helps avoid repeated calculations and improves performance within the given constraints.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not taking the modulo at each step. The intermediate calculations can become very large and overflow, leading to incorrect results. Remembering to apply modulo at each multiplication and addition is crucial.

✗Common mistake 2: Forgetting to subtract 1 from the final result. The problem asks for the number of *different* ways to reorder, so the original array should not be counted.

✗Common mistake 3: Not pre-computing the combinations table efficiently. Using direct calculation of nCr within the recursive function will result in TLE.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: In compiler design, similar logic can be used to analyze different parse trees that can be generated from a given grammar. Each parse tree is analogous to a different BST structure.

🏦Real-world scenario 2: In network routing, understanding the number of possible routing paths given certain constraints can be modeled similarly, where the constraints impose a tree-like structure on the routing options.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 8, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 14

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 7

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Number of Ways to Reorder Array to Get Same BST"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether math techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?