What is the time complexity of Cycle Length Queries in a Tree?

The optimal solution has a time complexity of O(q log n).

What data structures are used to solve Cycle Length Queries in a Tree?

This problem uses Array, Tree, Binary Tree. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Cycle Length Queries in a Tree?

Insight 1: The cycle is formed by the path from node 'a' to their lowest common ancestor (LCA), the path from node 'b' to the LCA, and the edge between 'a' and 'b'. The cycle length is the sum of these path lengths + 1. Insight 2: The problem can be solved by repeatedly dividing the larger of the two node values by 2 until they become equal. Each division represents moving one step up towards the root of the tree.

Which companies ask LeetCode 2509. Cycle Length Queries in a Tree in interviews?

This problem is frequently asked at Arcesium.

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Cycle Length Queries in a Tree - LeetCode 2509 Solution

Cycle Length Queries in a Tree - Complete Solution Guide

Cycle Length Queries in a Tree is LeetCode problem 2509, 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

You are given an integer n . There is a complete binary tree with 2 n - 1 nodes. The root of that tree is the node with the value 1 , and every node with a value val in the range [1, 2 n - 1 - 1] has two children where: The left node has the value 2 * val , and The right node has the value 2 * val + 1 . You are also given a 2D integer array queries of length m , where queries[i] = [a i , b i ] . For each query, solve the following problem: Add an edge between the nodes with values a i and b i .

Detailed Explanation

The problem describes a complete binary tree where nodes are numbered in a specific way: the root is 1, and a node with value 'val' has children 2*val and 2*val + 1. We're given a list of queries, each containing two node values (a, b). For each query, we need to imagine adding an edge between nodes 'a' and 'b', find the length of the resulting cycle, and then return that length. Importantly, we *don't* actually modify the tree; we only need to calculate the cycle length that *would* result. The key is understanding how node numbering relates to the tree structure and finding the shortest path between two nodes by navigating towards their common ancestor.

Solution Approach

The solution finds the cycle length for each query by calculating the combined path length from the two nodes in the query to their lowest common ancestor (LCA). It iteratively divides the larger of the two node values by 2, which simulates moving up the tree towards the root. This process continues until the two node values are equal, at which point the lowest common ancestor is found. The total number of steps taken during this process, plus 1, is the length of the cycle.

Step-by-Step Algorithm

Step 1: Initialize an empty list 'answer' to store the cycle lengths for each query.
Step 2: Iterate through each query (a, b) in the 'queries' list.
Step 3: Initialize 'path_len' to 0. This variable will store the combined length of the paths from 'a' and 'b' to their lowest common ancestor.
Step 4: Use a 'while' loop to continue as long as 'a' is not equal to 'b'.
Step 5: Inside the loop, compare 'a' and 'b'. If 'a' is greater than 'b', divide 'a' by 2 (integer division) and increment 'path_len'. This moves 'a' one level up the tree.
Step 6: If 'b' is greater than 'a', divide 'b' by 2 (integer division) and increment 'path_len'. This moves 'b' one level up the tree.
Step 7: Once the 'while' loop finishes (when 'a' equals 'b'), it means the lowest common ancestor has been found. The total path length is 'path_len'.
Step 8: Add 1 to 'path_len' to account for the direct edge between the original 'a' and 'b' nodes that completes the cycle. Append this sum to the 'answer' list.
Step 9: After processing all queries, return the 'answer' list.

Key Insights

Insight 1: The cycle is formed by the path from node 'a' to their lowest common ancestor (LCA), the path from node 'b' to the LCA, and the edge between 'a' and 'b'. The cycle length is the sum of these path lengths + 1.
Insight 2: The problem can be solved by repeatedly dividing the larger of the two node values by 2 until they become equal. Each division represents moving one step up towards the root of the tree.
Insight 3: The number of steps taken to reach the common ancestor, plus 1 (for the edge between the two nodes in the query), gives the cycle length.

Complexity Analysis

Time Complexity: O(q log n)

Space Complexity: O(q)

Topics

This problem involves: Array, Tree, Binary Tree.

Companies

Asked at: Arcesium.

Cycle Length Queries in a Tree | LeetCode 2509

Hard

Array Tree Binary Tree

Time: O(q log n)Space: O(q)

Solution Code

Available in:

class Solution {
    public int[] cycleLengthQueries(int n, int[][] queries) {
        int[] answer = new int[queries.length];
        for (int i = 0; i < queries.length; i++) {
            int a = queries[i][0];
            int b = queries[i][1];
            int path_len = 0;
            while (a != b) {
                if (a > b) {
                    a /= 2;
                } else {
                    b /= 2;
                }
                path_len += 1;
            }
            answer[i] = path_len + 1;
        }
        return answer;
    }
}

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Problem Info

Description

Problem 2509: Cycle Length Queries in a Tree - Detailed explanation and solution approaches.

Topics

Array Tree Binary Tree

Companies

Arcesium

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2509

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

9 steps

1
Step 1: Initialize an empty list 'answer' to store the cycle lengths for each query.
2
Step 2: Iterate through each query (a, b) in the 'queries' list.
3
Step 3: Initialize 'path_len' to 0. This variable will store the combined length of the paths from 'a' and 'b' to their lowest common ancestor.
4
Step 4: Use a 'while' loop to continue as long as 'a' is not equal to 'b'.
5
Step 5: Inside the loop, compare 'a' and 'b'. If 'a' is greater than 'b', divide 'a' by 2 (integer division) and increment 'path_len'. This moves 'a' one level up the tree.
6
Step 6: If 'b' is greater than 'a', divide 'b' by 2 (integer division) and increment 'path_len'. This moves 'b' one level up the tree.
7
Step 7: Once the 'while' loop finishes (when 'a' equals 'b'), it means the lowest common ancestor has been found. The total path length is 'path_len'.
8
Step 8: Add 1 to 'path_len' to account for the direct edge between the original 'a' and 'b' nodes that completes the cycle. Append this sum to the 'answer' list.
9
Step 9: After processing all queries, return the 'answer' list.

💡

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(q log n)

💾

Space Complexity

O(q)

📝

Why?

The time complexity depends on how many levels up the tree we need to traverse for each query to find the LCA. In the worst case, the two nodes are far apart in the tree, requiring us to traverse up to log n levels to find the LCA. Since we perform this operation for each of the q queries, the overall time complexity is O(q log n). The space complexity comes from storing the results of each query in the `answer` array, which grows linearly with the number of queries, resulting in O(q) space complexity.

📖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

Finding LCA using bit manipulation

Convert the node values to binary. Find the longest common prefix of the binary strings. The length of the cycle is then related to the lengths of the paths from the nodes to the node represented by the common prefix.

Time: O(q * log n) on average, but could be slightly better in some cases due to bitwise operations.Space: O(log n) for each query due to storing the binary representation of the node values.

Trade-offs: While potentially faster in certain scenarios due to efficient bitwise operations, this approach might be more complex to implement and understand, and the practical performance difference might not be significant compared to the given solution.

Precompute all the paths to root.

Before processing queries, precompute path from each node in the tree to the root. After that, compute the LCA from the two given nodes.

Time: O(n + q * k), where n is number of nodes in the binary tree (2^n-1) and k is the height of binary tree (log n).Space: O(n)

Trade-offs: It can be helpful to precompute and use that information for processing queries. But this approach can be helpful if the number of queries are very high.

✨

Key Insights

1Insight 1: The cycle is formed by the path from node 'a' to their lowest common ancestor (LCA), the path from node 'b' to the LCA, and the edge between 'a' and 'b'. The cycle length is the sum of these path lengths + 1.
2Insight 2: The problem can be solved by repeatedly dividing the larger of the two node values by 2 until they become equal. Each division represents moving one step up towards the root of the tree.
3Insight 3: The number of steps taken to reach the common ancestor, plus 1 (for the edge between the two nodes in the query), gives the cycle length.

📚

Key Terms Explained

For Students

Array

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

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: When a and b are siblings (e.g., 2 and 3). The algorithm correctly finds their parent (1) in one step each, resulting in path_len = 2, and the cycle length being 3.
Edge case 2: When a and b are on completely different branches of the tree. The algorithm will still correctly converge on their LCA (which will be 1 in this case).
Edge case 3: When a or b is 1. The algorithm will correctly determine the number of steps to the LCA (which is always 1). No special handling is needed due to the general nature of the iterative halving.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not accounting for the edge between nodes 'a' and 'b' when calculating the cycle length. The '+ 1' is crucial.

✗Common mistake 2: Using floating-point division instead of integer division (// in Python, / in Java and C++), which would lead to incorrect results due to the way nodes are numbered in the tree. Using the correct division operator (integer) is important for the algorithm's logic.

✗Common mistake 3: Incorrectly implementing the condition for moving up the tree. For example, not comparing 'a' and 'b' before dividing, leading to potential infinite loops or incorrect path lengths.

✅

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: Network routing. Finding the shortest path between two nodes in a network is a fundamental problem. This problem abstracts that idea to a binary tree structure, but the concept of finding the shortest path to a common ancestor is applicable.

🏦Real-world scenario 2: File system navigation. Imagine a file system where each directory is a node in a tree. Finding the common parent directory between two files is analogous to finding the LCA in this problem.

🎮Real-world scenario 3: Phylogenetic trees. In biology, phylogenetic trees represent the evolutionary relationships between different species. Finding the last common ancestor of two species can be done using algorithms similar to the one used here.

💡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

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ 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: November 17, 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 13

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

@DevExpert92Jul 6

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

@AlgoMasterJun 29

❓

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.

What is the difference between DFS and BFS for trees?

DFS (using recursion or stack) explores depth-first, useful for path-finding and backtracking problems. BFS (using queue) explores level-by-level, ideal for finding shortest paths or processing nodes by depth. Choose based on whether level information matters.

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 "Cycle Length Queries in a Tree"?

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 tree techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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